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(i) \(-5, 7\) Solution: Here, the given roots are \(5\) and \(7\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2-[(-5)+7]x+(-5)7=0\) \(\ ⇒ x^2-2x-35=0\) |
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(ii) \(-\frac{1}{2}, -3\) Solution: Here, the given roots are \(-\frac{1}{2}\) and \(-3\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2 -[(-\frac{1}{2})+(-3)]x+(-\frac{1}{2})(-3)=0 \) \(\ ⇒ x^2-(-\frac{7}{2})x +\frac{3}{2}=0 \) \(\ ⇒ x^2+\frac{7}{2} x+\frac{3}{2}=0 \) \(\ ⇒ 2x^2+7x+3=0 \) |
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(iii) \(1, -\frac{3}{2}\) Solution: Here, the given roots are \(1\) and \(-\frac{3}{2}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2-[(1)+(-\frac{3}{2})]x+(1)(-\frac{3}{2})=0 \) \(\ ⇒ x^2-(-\frac{1}{2})x-\frac{3}{2}=0 \) \(\ ⇒ x^2+\frac{1}{2}x-\frac{3}{2}=0 \) \(\ ⇒ 2x^2+x-3=0 \) |
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(iv) \(1, -\frac{4}{5}\) Solution: Here, the given roots are \(1\) and \(-\frac{4}{5}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2-[(1)+(-\frac{4}{5})]x+(1)(-\frac{4}{5})=0 \) \(\ ⇒ x^2-\frac{1}{5} x-\frac{4}{5}=0 \) \(\ ⇒ x^2-\frac{1}{5}x-\frac{4}{5}=0 \) \(\ ⇒ 5x^2-x-4=0 \) |
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(v) \(\frac{1}{2}, -\frac{1}{3}\) Solution: Here, the given roots are \(\frac{1}{2}\) and \(-\frac{1}{3}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2-[(\frac{1}{2})+(-\frac{1}{3})]x+(\frac{1}{2})(-\frac{1}{3})=0 \) \(\ ⇒ x^2-\frac{1}{6} x-\frac{1}{6}=0 \) \(\ ⇒ x^2-\frac{1}{6} x-\frac{1}{6}=0 \) \(\ ⇒ 6x^2-x-1=0 \) |
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(vi) \(5i, -5i\) Solution: Here, the given roots are \(5i\) and \(-5i\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \(\ ⇒ x^2-[(5i)+(-5i)]x+(5i)(-5i)=0 \) \(\ ⇒ x^2-0x-25i^2=0 \) \(\ ⇒ x^2-0-25(-1)=0 \) \(\ ⇒ x^2+25=0 \) |
2. Form the quadratic equation whose one root is
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(i) \( \sqrt{3}i \) Solution: Here, one root is \( \sqrt{3}i \) Therefore, other root will be \( - {\sqrt{3}i} \) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( ⇒ x^2-[(\sqrt{3}i)+(-\sqrt{3}i)]x+(\sqrt{3}i)(-\sqrt{3}i) = 0 \) \( ⇒ x^2-[0]x + [\sqrt{9} i^2] =0 \) \( ⇒ x^2-0 + [-3(-1)]=0\) \( ⇒ x^2+3=0 \) |
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(ii) \(4+\sqrt{5}\) Solution: Here, one root is \(4+\sqrt{5}\) ⸫ The other root will be \(4-\sqrt{5}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( ⇒ x^2- [(4+\sqrt{5})+(4-\sqrt{5})]x+[(4+\sqrt{5})(4-\sqrt{5})]=0 \) \( ⇒ x^2-[8]x+[(4)^2-(\sqrt{5})^2]=0 \) \( ⇒ x^2-8x+[16-5]=0 \) \( ⇒ x^2-8x+11=0 \) |
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(iii) \(\frac{1}{(2+\sqrt{3})}\) Solution: Here, one root is \( \frac{1}{(2+\sqrt{3})}=\frac{(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \) \(=\frac{(2-\sqrt{3})}{(2)^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \) ⸫ The root will be \(2+\sqrt{3}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( ⇒ x^2-[(2-\sqrt{3})+(2+\sqrt{3})]x+[(2-\sqrt{3})(2+\sqrt{3})]=0 \) \( ⇒ x^2-[4]x+[(2)^2-(\sqrt{3})^2]=0 \) \( ⇒ x^2-4x+[4-3]=0 \) \( ⇒ x^2-4x+1=0 \) |
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(iv) \(1-\sqrt{\frac{3}{2}}i\) Solution: Here, one root is \(1-\sqrt{\frac{3}{2}}i\) ⸫ The other root will be \(1+\sqrt{\frac{3}{2}}i\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( ⇒ x^2-[(1-\sqrt{\frac{3}{2}}i)+(1+\sqrt{\frac{3}{2}}i)]x+[(1-\sqrt{\frac{3}{2}}i)(1+\sqrt{\frac{3}{2}}i)]=0 \) \( ⇒ x^2-[2]x+[(1)^2-(\sqrt{\frac{3}{2}} i)^2]=0 \) \( ⇒ x^2-2x+[1-[\frac{3}{4 (-1)}] ]=0 \) \( ⇒ x^2-2x+[1+\frac{3}{4}]=0 \) \( ⇒ x^2-2x+\frac{7}{4}=0 \) \( ⇒ 4x^2-8x+7=0 \) |
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(v) \(\frac{-1-\sqrt{5}}{2}\) Solution: Here, one root is \(\frac{-1-\sqrt{5}}{2}\) ⸫ The other root will be \(\frac{-1+\sqrt{5}}{2}\) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( ⇒ x^2-[(\frac{-1-\sqrt{5}}{2})+(\frac{-1+\sqrt{5}}{2})]x+[(\frac{-1-\sqrt{5}}{2})(\frac{-1+\sqrt{5}}{2})]=0 \) \( ⇒ x^2-[\frac{-1-\sqrt{5}-1+\sqrt{5}}{2}] x+[\frac{(-1)^2-(\sqrt{5})^2}{4}]=0 \) \( ⇒ x^2-[\frac{-2}{2}] x+[\frac{1-5}{4}]=0 \) \( ⇒ x^2-x + [\frac{-4}{4}]=0 \) \( ⇒ x^2-x-1 =0 \) |
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(vi) \( p+\sqrt{p^2-c} \) Solution: Here, one root is \( p+\sqrt{p^2-c} \) ⸫ The other root will be \( p-\sqrt{p^2-c} \) ⸫ The required quadratic equation is, \(x^2-(sum~of~the~roots)x+(product~of~the~roots)=0\) \( x^2-[(p+\sqrt{p^2-c})+(p-\sqrt{p^2-c})] x\) \(+[(p+\sqrt{p^2-c})(p-\sqrt{p^2-c})]=0 \) \( ⇒ x^2-[2p]x+[(p)^2-(\sqrt{p^2-c})^2 ]=0 \) \( ⇒ x^2-[2p]x+[p^2-(p^2-c)]=0 \) \( ⇒ x^2-2px+[p^2-p^2+c]=0 \) \( ⇒ x^2-2px+p^2-p^2+c=0 \) \( ⇒ x^2-2px+c=0 \) |
3. Find a quadratic equation whose roots are
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(i) 2 less than the roots of \(\ x^2-16x+63=0 \) Solution: Let \(\ \alpha \) and \(\ \beta \) be the two roots of the given quadratic equation. Therefore, \(\alpha+\beta = 16 \) and \(\alpha \beta = 63 \) Let the new roots be \(\alpha-2 \) and \(\ \beta-2 \) ⸫ The required quadrartic equation is \(\ x^2 - \) (sum of the two new roots)\(\ x \) + (product of the two new roots) = 0 \(\ ⇒ x^2-[(\alpha-2)+(\beta-2)]x+[(\alpha-2)(\beta-2)]=0 \) \(\ ⇒ x^2-[(\alpha+\beta)-4]x+[\alpha\beta-2\alpha-2\beta+4]=0 \) \(\ ⇒ x^2-[16-4]x+[\alpha\beta-2(\alpha+\beta)+4]=0 \) \(\ ⇒ x^2-12x+[63-2(16)+4]=0 \) \(\ ⇒ x^2-12x+[67-32]=0 \) \(\ ⇒ x^2-12x+35=0 \) (Ans.) |
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(ii) Greater by 4 than the roots of the quadratic equation \(\ x^2+13x+5=0 \) Solution: Let \(\ \alpha \) and \(\ \beta \) be the two roots of the given quadratic equation. Therefore, \(\ \alpha+\beta = -13 \) and \(\ \alpha\beta = 5 \) Let the new roots be \(\ \alpha+4 \) and \(\ \beta+4 \) ⸫ The required quadrartic equation is \(\ x^2 - \) (sum of the two new roots)\(\ x \) + (product of the two new roots) = 0 \(\ ⇒ x^2-[(\alpha+4)+(\beta+4)]x+[(\alpha+4)(\beta+4)]=0 \) \(\ ⇒ x^2-[(\alpha+\beta)+8]x+[\alpha\beta+4\alpha+4\beta+16]=0 \) \(\ ⇒ x^2-[(-13)+8]x+[\alpha\beta+4(\alpha+\beta)+16]=0 \) \(\ ⇒ x^2-(-5)x+[5+4(-13)+16]=0 \) \(\ ⇒ x^2+5x+(21-52)=0 \) \(\ ⇒ x^2+5x-31=0 \) (Ans.) |
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(iii) Reciprocals of the roots of \(\ 2x^2-7x+6=0 \) Solution: Let \(\ \alpha \) and \(\ \beta \) be the two roots of the given quadratic equation. \(\alpha+\beta = \frac{7}{2} \) and \(\alpha \beta = \frac{6}{2} = 3 \) Let, the new roots be \(\ \frac{1}{\alpha} \) and \(\ \frac{1}{\beta} \) as they are reciprocals of the roots of the given quadratic equation. ⸫ The required quadrartic equation is \(\ x^2 - \) (sum of the two new roots)\(\ x \) + (product of the two new roots) = 0 \(\ ⇒ x^2- \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)x+ \frac{1}{\alpha} \frac{1}{\beta} = 0 \) \(\ ⇒ x^2 - \left(\frac{\beta+\alpha}{\alpha\beta}\right)x + \frac{1}{\alpha\beta} = 0 \) \(\ ⇒ x^2 - \left(\frac{\frac{7}{2}}{3}\right)x + \frac{1}{3} = 0 \) \(\ ⇒ x^2 - \left(\frac{7}{6}\right)x + \frac{1}{3} = 0 \) \(\ ⇒ 6x^2 - 7x +2 = 0 \) (Ans.) |
4. Find the value of \(\ k\) such that
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(i) one root of \( 2x^2-5x+k=0 \) is twice the other. Solution: Let \( \alpha \) be the one root of the quadratic equation \( 2x^2-5x+k=0 \) Then, the other root is \( 2\alpha \) Now, Sum of the two roots, \( \alpha+2\alpha = -\frac{(-5)}{2} \) \( ⇒ 3\alpha = \frac{5}{2} \) \( ⇒ \alpha = \frac{5}{6} \) ...(1) Also, Product of the two roots, \( \alpha ×2\alpha = \frac{k}{2} \) \( ⇒ 2\alpha^2 = \frac{k}{2} \) \( ⇒ 2\left(\frac{5}{6}\right) ^2 = \frac{k}{2} \) [using (1)] \( ⇒ 2×\frac{25}{36} = \frac{k}{2} \) \( ⇒ \frac{25}{18} = \frac{k}{2} \) \( ⇒ k = \frac{25}{9} \) |
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(ii) \( (2k-5)x^2-4x-15=0 \) and \( (3k-8)x^2-5x-21=0 \) have a common root. Solution: Let \( \alpha \) be the common roots of the given two quadratic equations, then \( (2k-5)\alpha^2-4\alpha-15=0 \) .....(1) \( (3k-8)\alpha^2-5\alpha-21=0 \) ......(2) On cross-multiplying we get, \( \frac{\alpha^2}{(-4)(-21)-(-15)(-5)}=\frac{\alpha}{(-15)(3k-8)-(2k-5)(-21)}=\frac{1}{(2k-5)(-5)-(-4)(3k-8)} \) ⇒ \( \frac{\alpha^2}{84-75}=\frac{\alpha}{-45k+120+42k-105}=\frac{1}{-10k+25+12k-32} \) ⇒ \( \frac{\alpha^2}{9}=\frac{\alpha}{-3k+15}=\frac{1}{2k-7} \) Now, \(\frac{\alpha^2}{9}=\frac{\alpha}{-3k+15}⇒\alpha=\frac{9}{-3k+15} \) ...(3) Again, \(\frac{\alpha}{-3k+15}=\frac{1}{2k-7} ⇒\alpha=\frac{-3k+15}{2k-7} \) ...(4) Comparing (3) and (4), we get \(\frac{9}{-3k+15}=\frac{-3k+15}{2k-7}\) ⇒ \(9(2k-7)=(-3k+15)^2 \) ⇒ \(18k-63=9k^2+225-90k \) ⇒ \(0=9k^2-90k-18k+225+63 \) ⇒ \(9k^2-108k+288=0 \) ⇒ \(9(k^2-12k+32)=0 \) ⇒ \(k^2-12k+32=0 \) [⸪ 9≠0] ⇒ \(k^2-8k-4k+32=0 \) ⇒ \(k(k-8)-4(k-8)=0 \) ⇒ \((k-8)(k-4)=0 \) ⇒ \(k=8\) or \(k=4\) |
5. Under what condition
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(i) \( 3x^2+4mx+2 =0 \) and \( 2x^2+3x-2=0 \) will have a common root? Solution: Let \( \alpha \) be the common roots between the given quadartic equations, then \( 3\alpha^2+4m\alpha+2 =0 \) and \( 2\alpha^2+3\alpha-2=0 \) On cross-multiplying, we get \( \frac{\alpha^2}{-8m-6}=\frac{\alpha}{4-(-6)}=\frac{1}{9-8m} \) \( = \frac{\alpha^2}{-8m-6}=\frac{\alpha}{4+6} ⇒ \alpha = \frac{-8m-6}{10} \) \( = \frac{\alpha}{4+6}=\frac{1}{9-8m} ⇒ \alpha = \frac{10}{9-8m} \) Comparing these two equations, we get \( \frac{-8m-6}{10} = \frac{10}{9-8m} \) \( ⇒ -72m+64m^2-54+48m = 100 \) \( ⇒ 64m^2-24m-154 = 0 \) \( ⇒ 2(32m^2-12m-77) = 0 \) \( ⇒ 32m^2-12m-77 = 0 \) \( ⇒ 32m^2-56m+44m-77 = 0 \) \( ⇒ 8m(4m-7)+11(4m-7) = 0 \) \( ⇒ (4m-7)(8m+11) = 0 \) Either \( 4m-7=0 ⇒ m = \frac{7}{4} \) Or \( 8m+11 = 0 ⇒ m = \frac{-11}{8} \) ⸫ The required condition for the given quadratic equations to have a common root is \( m = \frac{7}{4}, m = \frac{-11}{8} \) |
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(ii) one root of \(ax^2+bx+c=0\) will be \(n\) times the other? Solution: Let \(\alpha\) be one of the roots of the quadratic equation \(ax^2+bx+c=0\) and so the other root be \(n\alpha\) From the relation between roots and coefficients, we have, \(\alpha+n\alpha=-\frac{b}{a}\) ⇒ \(\alpha(1+n)=-\frac{b}{a}\) ⇒ \(\alpha=-\frac{b}{a(1+n)}\) ...(1) Again, \(n(n\alpha)=\frac{c}{a}\) ⇒ \(n\alpha^2=\frac{c}{a}\) ⇒ \(\alpha^2=\frac{c}{na}\) ⇒ \(\left[-\frac{b}{a(1+n)}\right]^2=\frac{c}{na}\) [using (1)] ⇒ \(\frac{b^2}{a^2(1+n)^2}=\frac{c}{na}\) ⇒ \(\frac{b^2}{ac}=\frac{(1+n)^2}{n}\) |
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(iii) one root of \(x^2-px+q=0\) will be twice the other. Solution: Let \(\alpha\) and \(2\alpha\) be the roots of the quadratic equation \(x^2-px+q=0\). From the relation between roots and coefficients, we have, \(\alpha+2\alpha=-(-p)\) ⇒ \(3\alpha=p\) ⇒ \(\alpha=\frac{p}{3}\) ...(1) Again, \(\alpha×2\alpha=q\) ⇒ \(2\alpha^2=q\) ⇒ \(\alpha^2=\frac{q}{2}\) ⇒ \(\left[\frac{p}{3}\right]^2=\frac{q}{2}\) [using (1)] ⇒ \(\frac{p^2}{9}=\frac{q}{2}\) ⇒ \(\frac{p^2}{q}=\frac{9}{2}\) |
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(iv) The roots of \(ax^2+bx+c=0\) will be in the ratio \(m:n\)? Solution: Let \(\alpha\) be the common ratio between the roots of the quadratic equation \(ax^2+bx+c=0\). So, the roots of the quadratic equation becomes \(m\alpha\) and \(n\alpha\). From the relation between roots and coefficients, we have, \(m\alpha+n\alpha-\frac{b}{a}\) ⇒ \(\alpha(m+n)=-\frac{b}{a}\) ⇒ \(\alpha=-\frac{b}{a(m+n)}\) ...(1) Again, \(m\alpha×n\alpha=\frac{c}{a}\) ⇒ \(mn\alpha^2=\frac{c}{a}\) ⇒ \(mn\left[-\frac{b}{a(m+n)}\right]^2=\frac{c}{a}\) [using (1)] ⇒ \(mn\left[\frac{b^2}{a^2(m+n)^2}\right]=\frac{c}{a}\) ⇒ \(mnb^2=ac(m+n)^2\) |
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(v) \(ax^2+bx+c=0\) and \(px^2+qx+r=0\) will have a common root? Solution: Let us consider two quadratic equations \( ax^2+bx+c=0 \)...(1) and \( px^2+qx+r=0 \) ...(2) Let \( \alpha \) be the common roots of the two quadratic equations, then \( a\alpha^2+b\alpha+c=0 \)...(3) and \( p\alpha^2+q\alpha+r=0 \) ...(4) On cross-multiplying, we get \( \frac{\alpha^2}{br-qc}=\frac{\alpha}{cp-ar}=\frac{1}{aq-bp} \) Now, \( \frac{\alpha^2}{br-qc}=\frac{\alpha}{cp-ar} \) \( ⇒ \alpha = \frac{br-qc}{cp-ar} ...(5) \) Also, \( \frac{\alpha}{cp-ar}=\frac{1}{aq-bp} \) \( ⇒ \alpha = \frac{cp-ar}{aq-bp} ...(6) \) Comparing (5) & (6), we get \( \frac{br-qc}{cp-ar} = \frac{cp-ar}{aq-bp} \) \( ⇒ (cp-ar)^2 = (br-qc)(aq-bp) \) This is the required condition for two quadratic equations to have one common root. |
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(vi) One root of \(ax^2+bx+c=0\) is four times the other? Solution: Let \(\alpha\) and \(4\alpha\) be the roots of the quadratic equation \(ax^2+bx+c=0\). From the relation between roots and coefficients, we have, \(\alpha+4\alpha-\frac{b}{a}\) ⇒ \(5\alpha=-\frac{b}{a}\) ⇒ \(\alpha=-\frac{b}{5a}\) ...(1) Again, \(m\alpha×4\alpha=\frac{c}{a}\) ⇒ \(4\alpha^2=\frac{c}{a}\) ⇒ \(4\left[-\frac{b}{5a}\right]^2=\frac{c}{a}\) [using (1)] ⇒ \(4\left[\frac{b^2}{25a^2}\right]=\frac{c}{a}\) ⇒ \(4b^2=25ac\) |
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(vii) the sum of the roots of \(x^2-mx+n=0\) is \(k\) times their difference? Solution: Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation \(x^2-mx+n=0\). From the relation between roots and coefficients, we have, \(\alpha+\beta=-(-m)=m\) Again, \(\alpha\beta=n\) According to question, \(\alpha+\beta=k(\alpha-\beta)\) ⇒ \((\alpha+\beta)^2=k^2(\alpha-\beta)^2\) [squaring both sides] ⇒ \((\alpha+\beta)^2=k^2\left[(\alpha+\beta)^2-4\alpha\beta\right] \) ⇒ \((\alpha+\beta)^2=k^2(\alpha+\beta)^2-4k^2\alpha\beta\) ⇒ \(4k^2\alpha\beta=k^2(\alpha+\beta)^2-(\alpha+\beta)^2\) ⇒ \(4k^2\alpha\beta=(\alpha+\beta)^2(k^2-1)\) ⇒ \(4k^2n=m^2(k^2-1)\) |
6. If \( \alpha \) and \( \beta \) are the roots of \( ax^2+bx+c =0 \) then express the value of the following symmetric function in terms of the co-efficients \( a, b, c \).
(i) \( \alpha^2+\alpha\beta+\beta^2 \)
(ii) \( (\alpha+2\beta)(2\alpha+\beta) \)
(iii) \( \alpha^4+\alpha^2\beta^2+\beta^4 \)
(iv) \( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} \)
Solution:
Since \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( ax^2+bx+c=0 \)
So, \( \alpha+\beta = \frac{-b}{a} \) and \( \alpha\beta=\frac{c}{a} \)
(i) \( \alpha^2+\alpha\beta+\beta^2 = \alpha^2+\beta^2+\alpha\beta \)
\( = (\alpha+\beta)^2-2\alpha\beta+\alpha\beta \)
\( = (\alpha+\beta)^2-\alpha\beta \)
\( = \frac{b^2}{a^2}-\frac{c}{a} \)
\( = \frac{b^2-ac}{a^2} \)
(ii) \( (\alpha+2\beta)(2\alpha+\beta)=2\alpha^2+\alpha\beta+4\alpha\beta+2\beta^2 \)
\( = 2(\alpha^2+\beta^2)+5\alpha\beta \)
\( = 2[(\alpha+\beta)^2-2\alpha\beta]+5\alpha\beta \)
\( = 2(\alpha+\beta)^2+\alpha\beta \)
\( = \frac{2b^2}{a^2}+\frac{c}{a} \)
\( = \frac{2b^2+ac}{a^2} \)
(iii) \( \alpha^4+\alpha^2\beta^2+\beta^4 = (\alpha^2)^2+(\beta^2)^2+(\alpha\beta)^2 \)
\( = (\alpha^2+\beta^2)^2-2\alpha^2\beta^2+(\alpha\beta)^2 \)
\( = [(\alpha^2+\beta^2)^2-2\alpha^2\beta^2]+(\alpha\beta)^2 \)
\( = (\alpha^2+\beta^2)^2-\alpha^2\beta^2 \)
\( = [(\alpha+\beta)^2-2\alpha\beta]^2 -(\alpha\beta)^2 \)
\( = [(\frac{-b}{a})^2-\frac{2c}{a}]^2-\frac{c^2}{a^2} \)
\( = (\frac{b^2}{a^2}-\frac{2c}{a})^2-\frac{c^2}{a^2} \)
\( = \frac{(b^2-2ac)^2}{a^4}-\frac{c^2}{a^2} \)
\( = \frac{(b^2-2ac)^2-c^2a^2}{a^4} \)
\( = \frac{b^4-4ab^2c+4a^2c^2-a^2c^2}{a^4} \)
\( = \frac{b^4-4ab^2c+3a^2c^2}{a^4} \)
(iv) \( \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} = \frac{\alpha^3+\beta^3}{\alpha\beta} \)
\( = \frac{(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)}{\alpha\beta} \)
\( = \frac{(\frac{-b}{a})^3-3\frac{c}{a}(\frac{-b}{a})}{\frac{c}{a}} \)
\( = \frac{\frac{-b^3}{a^3}+\frac{3bc}{a^2}}{\frac{c}{a}} \)
\( = \frac{\frac{-b^3+3abc}{a^3}}{\frac{c}{a}} \)
\( = \frac{-b^3+3abc}{a^2c} \)
7. If \( \alpha \) and \( \beta \) are the two roots of the quadratic equation \(ax^2+bx+c=0\) then find the quadratic equations having the following pairs of roots.
(i) \( \frac{\alpha}{\beta}, \frac{\beta}{\alpha} \)
(ii) \( \frac{1}{\alpha^2}, \frac{1}{\beta^2} \)
(iii) \( \alpha^4, \beta^4 \)
(iv) \( \sqrt{\frac{\alpha}{\beta}}, \sqrt{\frac{\beta}{\alpha}} \)
(v) \( \alpha^2+\beta^2, \frac{1}{\alpha^2}+\frac{1}{\beta^2} \)
(vi) \(\frac{1}{\alpha+\beta}, \frac{1}{\alpha}+\frac{1}{\beta}\)
(vii) \((\alpha-\beta)^2, (\alpha+\beta)^2\)
(viii) \(\alpha+2\alpha, \beta+2\alpha\)
(ix) \(\frac{\alpha^3}{\beta}, \frac{\beta^3}{\alpha}\)
(x) \(\alpha^2+\alpha\beta+\beta^2, \alpha^2-\alpha\beta+\beta^2\)
Solution:
⸪ \( \alpha \) and \( \beta \) are the two roots of the quadratic equation \( ax^2+bx+c = 0 \)
⸫ \( \alpha + \beta\ = \frac{-b}{a} \) and \( \alpha\beta = \frac{c}{a} \)
Now,
(i) Here the sum of the roots,
\( \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} \)
\( = \frac{[\alpha+\beta]^2 - 2\alpha\beta}{\alpha\beta} \)
\( = \frac{\frac{b^2}{a^2} - \frac{2c}{a}}{\frac{c}{a}} \)
\( = \frac{{\frac{b^2-2ac}{a^2}}}{\frac{c}{a}} \)
\( = \frac{b^2-2ac}{a^2}×\frac{a}{c} \)
\( = \frac{b^2-2ac}{ac} \)
and the product of the roots,
\( \frac{\alpha}{\beta}\frac{\beta}{\alpha} = 1 \)
⸫ The required quadratic equation is
\( x^2 - (\frac{\alpha}{\beta}+\frac{\beta)}{\alpha})x+ \frac{\alpha}{\beta}\frac{\beta}{\alpha} = 0 \)
\( ⇒ x^2 - (\frac{b^2-2ac}{ac})x+1 = 0 \)
\( ⇒ acx^2-(b^2-2ac)x+ac = 0 \)
(ii) Here the sum of the roots,
\( \frac{1}{\alpha^2}+\frac{1}{\beta^2} = \frac{\beta^2+\alpha^2}{\alpha^2\beta^2} \)
\( = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} \)
\( = \frac{(\frac{-b}{a})^2-2\frac{c}{a}}{(\frac{c}{a})^2} \)
\( = \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}} \)
\( = \frac{\frac{b^2-2ac}{a^2}}{\frac{c^2}{a^2}} \)
\( =\frac{b^2-2ac}{a^2}×\frac{a^2}{c^2} \)
\( = \frac{b^2-2ac}{c^2} \)
and the product of the roots,
\( \frac{1}{\alpha^2}×\frac{1}{\beta^2} = \frac{1}{(\alpha\beta)^2} \)
\( =\frac{1}{\left[\frac{c}{a}\right]^2}= \frac{1}{\frac{c^2}{a^2}} = \frac{a^2}{c^2} \)
⸫ The required quadratic equation is
\( x^2-[\frac{b^2-2ac}{c^2}]x + \frac{a^2}{c^2} = 0 \)
\( ⇒ c^2x^2-(b^2-2ac)x+a^2 = 0 \)
(iii) Here the sum of the roots,
\( \alpha^4+\beta^4 = (\alpha^2)^2+(\beta^2)^2 \)
\( = (\alpha^2+\beta^2)^2-2\alpha^2\beta^2 \)
\( = (\alpha^2+\beta^2)^2-2\alpha^2\beta^2 \)
\( = [(\alpha+\beta)^2-2\alpha\beta]^2 -2(\alpha\beta)^2 \)
\( = [(\frac{-b}{a})^2-\frac{2c}{a}]^2-2\left[\frac{c}{a}\right]^2 \)
\( = (\frac{b^2}{a^2}-\frac{2c}{a})^2-\frac{2c^2}{a^2} \)
\( = [\frac{b^2-2ac}{a^2}]^2-\frac{2c^2}{a^2} \)
\( = \frac{(b^2-2ac)^2}{a^4}-\frac{2c^2}{a^2} \)
\( = \frac{(b^2-2ac)^2-2a^2c^2}{a^4} \)
\( = \frac{(b^2)^2+(2ac)^2-2(b^2)(2ac)-2a^2c^2}{a^4} \)
\( = \frac{b^4++4a^2c^2-4ab^2c-2a^2c^2}{a^4} \)
\( = \frac{b^4+2a^2c^2-4ab^2c}{a^4} \)
and the product of the roots,
\( \alpha^4\beta^4 = (\alpha\beta)^4 \)
\( = \left[\frac{c}{a}\right]^4 = \frac{c^4}{a^4} \)
⸫ The required quadratic equation is
\( x^2-\left[\frac{b^4+2a^2c^2-4ab^2c}{a^4}\right]x+ \frac{c^4}{a^4} = 0 \)
\( ⇒ a^4x^2-(b^4-4ab^2c+2a^2c^2)x+c^4 = 0 \)
(iv) Here the sum of the roots,
\( \sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}}+\frac{\sqrt{\beta}}{\sqrt{\alpha}} \)
\( = \frac{\alpha+\beta}{\sqrt{\alpha\beta}} \)
\( = \frac{\frac{-b}{a}}{\sqrt{\frac{c}{a}}} \)
\( = \frac{-b}{a}×\sqrt{\frac{a}{c}} \)
\( = \frac{-b}{a}×\frac{\sqrt{a}}{\sqrt{c}} \)
\( = \frac{-b}{\sqrt{a}\sqrt{a}}×\frac{\sqrt{a}}{\sqrt{c}} \)
\( = \frac{-b}{\sqrt{a}×\sqrt{c}} \)
\( = \frac{-b}{\sqrt{ac}} \)
and the product of the roots,
\( \sqrt{\frac{\alpha}{\beta}}×\sqrt{\frac{\beta}{\alpha}} = \frac{\sqrt{\alpha}}{\sqrt{\beta}}×\frac{\sqrt{\beta}}{\sqrt{\alpha}} = 1 \)
⸫ The required quadratic equation is
\( x^2-\left[\frac{-b}{\sqrt{ac}}\right]x + 1 = 0 \)
\( ⇒ x^2+\frac{b}{\sqrt{ac}}x+1 = 0 \)
\( ⇒ \sqrt{ac}x^2+bx+\sqrt{ac} = 0 \)
(v) Here the sum of the roots,
\( (\alpha^2+\beta^2)+\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right) = (\alpha^2+\beta^2)+\left(\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\right) \)
\(=(\alpha^2+\beta^2)\left(1+\frac{1}{\alpha^2\beta^2}\right) \)
\(= [(\alpha)+\beta)^2-2\alpha\beta]\left[1+\frac{1}{(\alpha\beta)^2}\right] \)
\(=\left[\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right] \left[1+\frac{1}{\left(\frac{c}{a}\right)^2}\right] \)
\(=\left(\frac{b^2}{a^2}-\frac{2c}{a}\right)\left(1+\frac{a^2}{c^2}\right) \)
\(=\left(\frac{b^2-2ac}{a^2}\right)\left(\frac{c^2+a^2}{c^2}\right) \)
\(=\frac{(b^2-2ac)(a^2+c^2)}{a^2c^2} \)
and the product of the roots,
\( (\alpha^2+\beta^2×\left(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\right) = (\alpha^2+\beta^2)×\left(\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\right) \)
\(=\frac{(\alpha^2+\beta)^2}{\alpha^2\beta^2} \)
\(=\frac{\left[(\alpha+\beta)^2-2\alpha\beta\right]^2}{(\alpha\beta)^2} \)
\(=\frac{\left[\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right]^2}{\left(\frac{c}{a}\right)^2} \)
\(=\frac{\left(\frac{b^2}{a^2}-\frac{2c}{a}\right)^2}{\frac{c^2}{a^2}} \)
\(=\frac{\left(\frac{b^2-2ac}{a^2}\right)^2}{\frac{c^2}{a^2}} \)
\(=\left(\frac{b^2-2ac}{a^2}\right)^2×\frac{a^2}{c^2} \)
\(=\frac{(b^2-2ac)^2}{a^4}×\frac{a^2}{c^2} \)
\(= \frac{(b^2-2ac)^2}{a^2c^2}\)
⸫ The required quadratic equation is
\( x^2-\left[\frac{(b^2-2ac)(a^2+c^2)}{a^2c^2}\right]x + \frac{(b^2-2ac)^2}{a^2c^2} = 0 \)
\( ⇒ ac^2x^2-(b^2-2ac)(a^2+c^2)x+(b^2-2ac)^2= 0 \)
(vi) Here the sum of the roots,
\( \frac{1}{\alpha+\beta}+\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{1}{\alpha+\beta}+\frac{\beta+\alpha}{\alpha\beta}\)
\(=\frac{1}{\frac{-b}{a}}+\frac{\frac{-b}{a}}{\frac{c}{a}} \)
\(=\frac{-a}{b}+\left(\frac{-b}{a}\right)×\frac{a}{c} \)
\(=-\frac{a}{b}-\frac{b}{c}= -\left(\frac{a}{b}+\frac{b}{c}\right)\)
\(=-\left(\frac{ac+b^2}{bc}\right)\)
and the product of the roots,
\( \frac{1}{\alpha+\beta}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)=\frac{1}{\alpha+\beta}\left(\frac{\beta+\alpha}{\alpha\beta}\right)\)
\(=\frac{1}{\alpha\beta}=\frac{1}{\frac{c}{a}}=\frac{a}{c} \)
⸫ The required quadratic equation is
\( x^2-\left[-\left(\frac{ac+b^2}{bc}\right)\right]x+\frac{a}{c} = 0 \)
\( ⇒ bcx^2+(ac+b^2)x+ab= 0 \)
(vii) Here the sum of the roots,
\((\alpha-\beta)^2+(\alpha+\beta)^2=2(\alpha^2+\beta^2) \)
\(=2[(\alpha+\beta)^2-2\alpha\beta] \)
\(=2\left[\left(\frac{-b}{a}\right)^2-2\left(\frac{c}{a}\right)\right] \)
\(=2\left[\frac{b^2}{a^2}-\frac{2c}{a}\right] =2\left(\frac{b^2-2ac}{a^2}\right)\)
and the product of the roots,
\( (\alpha-\beta)^2×(\alpha+\beta)^2=[(\alpha+\beta)^2-4\alpha\beta]×(\alpha+\beta)^2 \)
\(=\left[\left(\frac{-b}{a}\right)^2-4\left(\frac{c}{a}\right)\right] \left[\frac{-b}{a}\right]\)
\(=\left[\frac{b^2}{a^2}-\frac{4c}{a}\right] \left[\frac{b^2}{a^2}\right]\)
\(=\left(\frac{b^2-4ac}{a^2}\right) \left(\frac{b^2}{a^2}\right)=\frac{b^2(b^2-4ac)}{a^4}\)
⸫ The required quadratic equation is
\(x^2-\left[2\left(\frac{b^2-2ac}{a^2}\right)\right]x+\frac{b^2(b^2-4ac)}{a^4}=0 \)
\( ⇒a^4x^2-2a^2(b^2-2ac)x+b^2(b^2-4ac)=0\)
(viii) Here the sum of the roots,
\((\alpha+2\alpha)+(\beta+2\alpha)=\alpha+2\alpha+\beta+2\alpha\)
\(=3(\alpha+\beta)=3\left(\frac{-b}{a}\right)=-\frac{3b}{a}\)
and the product of the roots,
\((\alpha+2\alpha)(\beta+2\alpha)=\alpha\beta+2\alpha^2+2\beta^2+4\alpha\beta\)
\(=2(\alpha^2+\beta^2)+5\alpha\beta \)
\(=2\left[(\alpha+\beta)^2-2\alpha\beta\right]+5\alpha\beta \)
\(=2(\alpha+\beta)^2-4\alpha\beta+5\alpha\beta=2(\alpha+\beta)^2+\alpha\beta\)
\(=2\left(\frac{-b}{a}\right)^2+\frac{c}{a}=\frac{2b^2}{a^2}+\frac{c}{a}=\frac{2b^2+ac}{a^2}\)
⸫ The required quadratic equation is
\( x^2-\left(-\frac{3b}{a}\right)x+\frac{2b^2+ac}{a^2}=0 \)
\( ⇒ a^2x^2+3abx+2b^2+ac=0\)
(ix) Here the sum of the roots,
\( \frac{\alpha^3}{\beta}+\frac{\beta^3}{\alpha}=\frac{\alpha^4+\beta^4}{\alpha\beta}\)
\(=\frac{(\alpha^2)^2+(\beta^2)^2}{\alpha\beta}=\frac{(\alpha^2+\beta^2)^2-2\alpha^2\beta^2}{\alpha\beta} \)
\(=\frac{[(\alpha+\beta)^2-2\alpha\beta]^2-2(\alpha\beta)^2}{\alpha\beta} \)
\(=\frac{\left[\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right]^2-2\left(\frac{c}{a}\right)^2}{\frac{c}{a}} \)
\(=\frac{\left[\frac{b^2}{a^2}-\frac{2c}{a}\right]^2-\frac{2c^2}{a^2}}{\frac{c}{a}} \)
\(=\frac{\left[\frac{b^2-2ac}{a^2}\right]^2-\frac{2c^2}{a^2}}{\frac{c}{a}} \)
\(=\frac{\frac{b^4+4a^2c^2-4ab^2c}{a^4}-\frac{2c^2}{a^2}}{\frac{c}{a}} \)
\(=\frac{b^4+4a^2c^2-4ab^2c-2a^2c^2}{a^4}×\frac{c}{a} \)
\(=\frac{b^4+2a^2c^2+4ab^2c}{a^3c} \)
and the product of the roots,
\( \frac{\alpha^3}{\beta}\frac{\beta^3}{\alpha}=\alpha^2\beta^2=(\alpha\beta)^2\)
\(=\left(\frac{c}{a}\right)^2=\frac{c^2}{a^2} \)
⸫ The required quadratic equation is
\(x^2-\left(\frac{b^4+2a^2c^2+4ab^2c}{a^3c}\right)x+\left(\frac{c}{a}\right)^2+\frac{c^2}{a^2}\)
\( ⇒a^3cx^2-(b^4+2a^2c^2-4ab^2c)x+ac^3=0\)
(x) Here the sum of the roots,
\( \alpha^2+\alpha\beta+\beta^2+\alpha^2-\alpha\beta+\beta^2=2(\alpha^2+\beta^2) \)
\(=2[(\alpha+\beta)^2-2\alpha\beta] \)
\(=2\left[\left(-\frac{b}{a}\right)^2-2\frac{c}{a}\right] \)
\(=2\left[\frac{b^2}{a^2}-\frac{2c}{a}\right]\)
\(=2\left(\frac{b^2-2ac}{a^2}\right)=\frac{2b^2-4ac}{a^2}\)
and the product of the roots,
\((\alpha^2+\alpha\beta+\beta^2)(\alpha^2-\alpha\beta+\beta^2) \)
\(=(\alpha^2+\beta^2+\alpha\beta)(\alpha^2+\beta^2-\alpha\beta) \)
\(=(\alpha+\beta)^2-(\alpha\beta)^2 \)
\(=[(\alpha+\beta)^2-2\alpha\beta]^2-(\alpha\beta)^2 \)
\(=\left[\left(-\frac{b}{a}\right)^2-2\left(\frac{c}{a}\right)\right]^2-\left(\frac{c}{a}\right)^2 \)
\(=\left[\frac{b^2}{a^2}-\frac{2c}{a}\right]^2-\frac{c^2}{a^2} \)
\(=\left(\frac{b^2-2ac}{a^2}\right)^2-\frac{c^2}{a^2} \)
\(=\frac{b^4+4a^2c^2-4ab^2c}{a^4}-\frac{c^2}{a^2} \)
\(=\frac{b^4+4a^2c^2-4ab^2c-a^2c^2}{a^4} \)
\(=\frac{b^4+3a^2c^2-4ab^2c}{a^4} \)
⸫ The required quadratic equation is
\( x^2-\left(\frac{2b^2-4ac}{a^2}\right)x+\frac{b^4+3a^2c^2-4ab^2c}{a^4}=0\)
\( ⇒a^4x^2-(2a^2b^2-4a^3c)x+(b^4+3a^2c^2-4ab^2c)=0\)
8. If \( a^2 = 5a - 3 \) and \( b^2 = 5b - 3 (a ≠ b) \), then find the quadratic equation whose roots are \( \frac{a}{b} \) and \( \frac{b}{a} \).
Solution:
Here, \( a^2=5a-3 ⇒ a^2-5a+3=0 \)
and \( b^2=5a-3 ⇒ b^2-5a+3=0 \)
⸫\( x^2 - 5x + 3 = 0 \) is the two quadratic equation whose roots are \( a \) and \( b \)
⸫\( a + b = 5 \) and \( ab = 3 \)
Now, \( \frac{a}{b}+\frac{b}{a} = \frac{a^2+b^2}{ab} = \frac{(a+b)^2-2ab}{ab} \)
\( = \frac{5^2-2(3)}{3} = \frac{19}{3} \)
Also, \( \frac{a}{b}\frac{b}{a} = 1 \)
⸫ The required quadratic equation having the roots \( \frac{a}{b} \) and \( \frac{b}{a} \) is, \( x^2-\left(\frac{19}{3}\right)x +1 = 0 ⇒ 3x^2 -19x+3 = 0 \)
9. If \( p\) and \( q\) are the roots of \( 3x^2 + 6x + 2 = 0\), then find the quadratic equation having the roots \( -\frac{p^2}{q}, -\frac{q^2}{p} \)
Solution:
Given, \( p\) and \( q\) are the roots of \( 3x^2 + 6x + 2 = 0\)
⸫ \( p + q = -\frac{6}{3} = -2 \) and \( ab = \frac{2}{3} \)
Now, \( \left(-\frac{p^2}{q}\right) + \left(-\frac{q^2}{p}\right) = \frac{-p^3-q^3}{pq} \)
\( = \frac{-(p^3+q^3)}{pq} = \frac{-[(p+q)^3-3pq(p+q)]}{pq} \)
\( = \frac{-[(-2)^3-3(\frac{2}{3})(-2)]}{\frac{2}{3}} \)
\( = \frac{-(-8+4)}{\frac{2}{3}} = \frac{4 ×3}{2} = 6 \)
Also, \( \left(-\frac{p^2}{q}\right) ×\left(-\frac{q^2}{p}\right) = pq = \frac{2}{3} \)
⸫ The required quadratic equation having the roots \( -\frac{p^2}{q}, -\frac{q^2}{p} \) is, \( x^2 - 6x + \frac{2}{3} = 0 ⇒ 3x^2 - 18x + 2 = 0 \)
10. If 4 is \( a\) root of \( x^2 + ax + 8 = 0\) and the roots of \( x^2 + ax + b = 0\) are equal, then find the value of \( b\).
Solution:
⸪ 4 is a root of the quadratic equations \( x^2 + ax + 8 = 0\)
⸫ \( (4)^2+4a+8 = 0 \)
\( ⇒ 16+4a+8 = 0 \)
\( ⇒ 4a = -24 \)
\( ⇒ a = -6 \) ...(1)
⸪ The quadratic equation \( x^2 + ax + b = 0\) is having same roots.
⸫ \( b^2 - 4ac = 0 \)
⸫ \( a^2-4(1)(b)=0 \)
\( ⇒ (-6)^2-4b = 0 \) [using (1)]
\( ⇒ 36 - 4b = 0 \)
\( ⇒ -4b = -36 \)
\( ⇒ b = 9 \)
11. If \( \alpha \) be one root of \( 4x^2 + 2x -1 = 0\), then show that the other root is \( 4\alpha^3-3\alpha \)
Solution:
⸪ \( \alpha \) is one root of the quadratic equation \( 4x^2 + 2x -1 = 0\)
⸫ \( 4\alpha^2+2x-1 - 0 ⇒ 4\alpha^2 = 1-2\alpha \) ...(1)
\( ⇒ \alpha^2 = \frac{1-2\alpha}{4} \) ...(2)
Also, we have \( 4\alpha^3-3\alpha \) as the other root of the given quadratic equation. Let us now reduce its degree,
\( 4\alpha^3-3\alpha = 4\alpha^2.\alpha - 3\alpha \)
\( = \alpha(1-2\alpha) - 3\alpha \) [using (1)]
\( = \alpha-2\alpha^2-3\alpha \)
\( = -2\alpha^2-2\alpha \)
\( = -2(\frac{1-2\alpha}{4})-2\alpha \) [using (2)]
\( = -\frac{1}{2}+\alpha-2\alpha \)
\( = -\alpha-\frac{1}{2} \)
\( ⇒ 4\alpha^3-3\alpha = -(\alpha+\frac{1}{2}) \)
Now, we will see if the reduced value of the root other than \( \alpha \) satisfies the given quadratic equation,
LHS= \( 4[-(\alpha+\frac{1}{2})]^2 + 2[-(\alpha+\frac{1}{2})] - 1 \)
\( = 4(\alpha^2+\alpha+\frac{1}{4}) + 2\alpha - 1 - 1 \)
\( = 4\alpha^2 + 1 + 4\alpha -2\alpha - 2 \)
\( = 1-2\alpha + 1 + 4\alpha - 2\alpha - 2 \) [using (1)]
\( = 0 \) = RHS
⸫ When \( \alpha \) is one root of \( 4x^2 + 2x -1 = 0\) then the other root is \( 4\alpha^3-3\alpha \)
Hence shown.
12. If the difference of the two roots of \( x^2 + px + q = 0\) is 1, then show that \( p^2 + 4q^2 = (1 + 2q)^2 \)
Solution:
⸪ The difference between the two roots of the given quadratic equation is 1, let us consider \(\ \alpha \) and \(\ \alpha-1 \) are the roots of the quadratic equation \( x^2 + px + q = 0\)
⸫ \(\ \alpha + \alpha - 1 = -p \)
\(\ ⇒ 2\alpha - 1 = -p ⇒ \alpha = \frac{1-p}{2} \) ...(1)
Also, \(\ \alpha (\alpha - 1) = q \)
\(\ ⇒ \left(\frac{1-p}{2}\right) \left(\frac{1-p}{2}-1\right)= q \) [using (1)]
\(\ ⇒\left(\frac{1-p}{2}\right) \left(\frac{1-p-2}{2}\right) = q \)
\(\ ⇒ \left[\frac{(1-p)}{2}\right] \left[\frac{(-1-p)}{2}\right] = q \)
\(\ ⇒ -\frac{(1-p)}{2}\frac{(1+p)}{2} = q \)
\(\ ⇒ -\frac{1+p^2}{4} = q \)
\(\ ⇒ p^2-1 = 4q \)
\(\ ⇒ p^2 = 4q + 1 ...(2) \)
Now, LHS = \( p^2+4q^2 = (4q+1)+ 4q^2 \) [using (2)]
\(\ = 1 + 4q + 4q^2 = (1+2q)^2 \)= RHS
Hence shown.
13. If \(\ ax^2 + bx + c = 0 \) and \(\ bx^2 + cx + a = 0 \) have a common root, then prove that \(\ a + b + c = 0 \) or \(\ a = b = c \) .
Solution:
Let \( \alpha \) be the common roots of the two given quadratic equations.
⸫ \( a\alpha^2+b\alpha+c=0 \) ...(1)
\( b\alpha^2+c\alpha+a=0 \) ...(2)
On cross-multiplying, we get,
\( \frac{\alpha^2}{ab-c^2} = \frac{\alpha}{bc -a^2} = \frac{1}{ac-b^2} \)
⸫ \( \frac{\alpha^2}{ab-c^2} = \frac{\alpha}{bc-a^2} ⇒ \alpha = \frac{ab-c^2}{bc-a^2} \) ...(3)
Also, \( \frac{\alpha}{bc-a^2} = \frac{a}{ac-b^2} ⇒ \alpha = \frac{bc-a^2}{ac-b^2} \) ...(4)
Comparing (3) & (4), we get
\( \frac{ab-c^2}{bc-a^2} = \frac{bc-a^2}{ac-b^2} \)
\( ⇒ (ab-c^2)(ac-b^2) = (bc-a^2)^2 \)
\( ⇒ a^2bc-ab^3-ac^3+b^2c^2 = b^2c^2+a^4-2a^2bc \)
\( ⇒ a^2bc+2a^2bc-ab^3-ac^3-a^4 = 0 \)
\( ⇒ -a^4-ab^3-ac^3+3a^2bc = 0 \)
\( ⇒ -a(a^3+b^3+c^3-3abc) = 0 \)
\( ⇒ a^3+b^3+c^3-3abc = 0 [⸪ -a≠0] \)
\( ⇒ \frac{1}{2}(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2] = 0 \)
\( ⇒ (a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2] = 0\) [⸪ \(\frac{1}{2} ≠ 0\)]
⸫ Either a+b+c = 0
or \( (a-b)^2+(b-c)^2+(c-a)^2 = 0 \)
Now, for the second case, since the sum of three positive term is zero, so, each term must be equal to zero. Therefore,
\( (a-b)^2 = 0 ⇒ a-b = 0 ⇒ a = b \)
\( (b-c)^2 = 0 ⇒ b-c = 0 ⇒ b = c \)
\( (c-a)^2 = 0 ⇒ c-a = 0 ⇒ c = a \)
⸫ \(\ a = b = c \)
Hence proved.
14. If the two roots of \(\ ax^2 + bx + a = 0 \) are equal, then show that \( \frac{a^2+b^2}{a^2-b^2}=-\frac{5}{3} \)
Solution:
⸪ The two roots of \(\ ax^2 + bx + a = 0 \) are equal,
⸫ \(\ b^2 - 4ac = 0 \)
\( ⇒ b^2-4a^2=0 \) [⸪ \(\ c=a \)]
\( ⇒ b^2 = 4a^2 \) ...(1)
Now, \( \frac{a^2+b^2}{a^2-b^2} = \frac{a^2+4a^2}{a^2-4a^2} \) [using (1)]
\( = \frac{5a^2}{-3a^2} \)
\( ⇒ \frac{a^2+b^2}{a^2-b^2} = -\frac{5}{3} \)
Hence shown.
15. Solve
(i) \( x^4-13x^2+36=0 ...(1) \)
Let \( x^2=y \)
⸫ (1) ⇒\( y^2-13y+36 = 0 \)
\( ⇒ y^2-(9+4)y+36 = 0 \)
\( ⇒ y^2-9y-4y+36 = 0 \)
\( ⇒ y(y-9)-4(y-9) = 0 \)
\( ⇒ (y-9)(y-4) = 0 \)
Either \( y = 9 \)
\( ⇒ x^2=9 \)
\( ⇒ x = \pm 3 \)
or \( y = 4 \)
\( ⇒ x^2=4 \)
\( ⇒ x = \pm 2 \)
⸫ The required roots are 3, -3, 2 and -2
(ii) \( x^4-3x^2+2 = 0 ...(1) \)
Let \( x^2=y \)
⸫ (1) ⇒\( y^2-3y+2 = 0 \)
\( ⇒ y^2-(2+1)y+2 = 0 \)
\( ⇒ y^2-2y-y+2 = 0 \)
\( ⇒ y(y-2)-1(y-2) = 0 \)
\( ⇒ (y-2)(y-1) = 0 \)
Either \( y=2 ⇒ x^2 = 2 \)
\( ⇒ x = \pm \sqrt{2} \)
or \( y=1 ⇒ x^2 = 1 \)
\( ⇒ x = \pm 1 \)
⸫ The required roots are \( \sqrt{2}, -\sqrt{2}, 1 \) and \( -1 \)
(iii) \( (x^2-3x)^2-5(x^2-3x)+6 = 0 ...(1) \)
Let \( x^3-3x=y \)
⸫ (1) ⇒ \( y^2-5y+6 = 0 \)
\( ⇒ y^2-(2+3)y+6 = 0 \)
\( ⇒ y^2-2y-3y+6 = 0 \)
\( ⇒ y(y-2)-3(y-2) = 0 \)
\( ⇒ (y-2)(y-3) = 0 \)
Either \( y-2 = 0 \)
\( ⇒ x^2-3x-2 = 0 \)
\( ⇒ x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-2)}}{2(1)} \)
\( ⇒ x = \frac{3\pm\sqrt{9+8}}{2} \)
\( ⇒ x = \frac{3\pm \sqrt{17}}{2} \)
or \( y-3 = 0 \)
\( ⇒ x^2-3x-3 = 0 \)
\( ⇒ x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-3)}}{2(1)} \)
\( ⇒ x = \frac{3\pm\sqrt{9+12}}{2} \)
\( ⇒ x = \frac{3\pm \sqrt{21}}{2} \)
⸫ The required roots are \( \frac{3\pm \sqrt{17}}{2}, \frac{3\pm \sqrt{21}}{2} \)
(iv) \( (x^2+2x-3)^2-3(x^2+2x-1)+8 = 0....(1) \)
Let \( x^2+2x-1 = y \)
⸫ (1) ⇒ \( (y-2)^2-3y+8 = 0 \)
\( ⇒ y^2-2(y)(2)+4-3y+8 = 0 \)
\( ⇒ y^2-4y-3y+8 = 0 \)
\( ⇒ y^2-7y+12 = 0 \)
\( ⇒ y^2-(4+3)y+12 = 0 \)
\( ⇒ y^2-4y-3y+12 = 0 \)
\( ⇒ y(y-4)-3(y-4) = 0 \)
\( ⇒ (y-4)(y-3) = 0 \)
Either \( y-4=0 \)
\( ⇒ x^2+2x-1-4 = 0 \)
\( ⇒ x^2+2x-5 = 0 \)
\( ⇒ x = \frac{-(2) \pm \sqrt{(2)^2-4(1)(-5)}}{2(1)} \)
\( ⇒ x = \frac{-2\pm\sqrt{4+20}}{2} \)
\( ⇒ x = \frac{-2\pm \sqrt{24}}{2} \)
\( ⇒ x = \frac{-2\pm 2\sqrt{6}}{2} \)
or \( y-3=0 \)
\( ⇒ x^2+2x-1-3 = 0 \)
\( ⇒ x^2+2x-4 = 0 \)
\( ⇒ x = \frac{-(2) \pm \sqrt{(2)^2-4(1)(-4)}}{2(1)} \)
\( ⇒ x = \frac{-2\pm\sqrt{4+16}}{2} \)
\( ⇒ x = \frac{-2\pm \sqrt{20}}{2} \)
\( ⇒ x = \frac{-2\pm 2\sqrt{5}}{2} \)
⸫ The required roots are \( \frac{-2\pm 2\sqrt{6}}{2}, \frac{-2\pm 2\sqrt{5}}{2} \)
(v) \( x^2-5x+10 = 5\sqrt{x^2-5x+4}...(1) \)
Let \( y = x^2-5x+4 \)
⸫ (1) ⇒ \( (y+6) = 5\sqrt{y} \)
\( ⇒ (y+6)^2 = (5\sqrt{y})^2 \)
\( ⇒ y^2+2(y)(6)+6^2 = 5^2y \)
\( ⇒ y^2+12y+36 = 25y \)
\( ⇒ y^2-13y+36 = 0 \)
\( ⇒ y^2-9y-4y+36 = 0 \)
\( ⇒ y(y-9)-4(y-9) = 0 \)
\( ⇒ (y-9)(y-4) = 0 \)
Either \( y-9 = 4 \)
\( ⇒ x^2-5x+4-9 = 0 \)
\( ⇒ x^2-5x-5 = 0 \)
\( ⇒ x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-5)}}{2(1)} \)
\( ⇒ x = \frac{5\pm\sqrt{25+20}}{2} \)
\( ⇒ x = \frac{5\pm \sqrt{45}}{2} \)
\( ⇒ x = \frac{5\pm 3\sqrt{5}}{2} \)
or \( y-4=0 \)
\( ⇒ x^2-5x+4-4 = 0 \)
\( ⇒ x^2-5x = 0 \)
\( ⇒ x(x-5) = 0 \)
\( ⇒ x = 0 \)
or \( x = 5 \)
⸫ The required roots are 0, 5 and \(\frac{5\pm 3\sqrt{5}}{2} \)
Alternative Solution:
\( x^2-5x+10 = 5\sqrt{x^2-5x+4} \)
\( ⇒ x^2-5x+10-5\sqrt{x^2-5x+4} = 0 ...(1) \)
Let \( y = \sqrt{x^2-5x+4} \)
⸫ (1) ⇒ \( y^2 = x^2-5x+4 \)
\( ⇒ y^2-4 = x^2-5x \)
\( (1)⇒ y^2-4+10-5y = 0 \)
\( ⇒ y^2-5y+6 = 0 \)
\( ⇒ y^2-2y-3y+6 = 0 \)
\( ⇒ y(y-2)-3(y-2) = 0 \)
\( ⇒ (y-2)(y-3) =0 \)
Either \( y^2 = 9 \)
\( ⇒ x^2-5x+4 = 9 \)
\( ⇒ x^2-5x-5=0 \)
\( ⇒ x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-5)}}{2(1)} \)
\( ⇒ x = \frac{5\pm\sqrt{25+20}}{2} \)
\( ⇒ x = \frac{5\pm \sqrt{45}}{2} \)
\( ⇒ x = \frac{5\pm 3\sqrt{5}}{2} \)
or \( y^2 = 4 \)
\( ⇒ x^2-5x-4 = 4 \)
\( ⇒ x^2-5x = 0 \)
\( ⇒ x(x-5) = 0 \)
\( ⇒ x = 0 \) or \( x = 5 \)
⸫ The required roots are 0, 5 and \(\frac{5\pm 3\sqrt{5}}{2} \)
(vi) \( \sqrt{x^2+5x-2}+\sqrt{x^2+5x-3}...(1) \)
Let \( x^2+5x-2 = y \)
⸫ (1) ⇒ \( \sqrt{y}+\sqrt{y-3} = 3 \)
\( ⇒ \sqrt{y} = 3-\sqrt{y-3} \)
\( ⇒ (\sqrt{y})^2 = (3-\sqrt{y-3})^2 \)
\( ⇒ y = 3^2-2(3)(\sqrt{y-3})+(\sqrt{y-3})^2 \)
\( ⇒ y = 9-6\sqrt{y-3}+y-3 \)
\( ⇒ -6 = -6\sqrt{y-3} \)
\( ⇒ 1 = \sqrt{y-3} \)
\( ⇒ 1 = y-3 \)
\( ⇒ y = 4 \)
\( ⇒ x^2+5x-2 = 4 \)
\( ⇒ x^2+5x-6 = 0 \)
\( ⇒ x^2 +(6-1)x-6 = 0 \)
\( ⇒ x^2+6x-x-6 = 0 \)
\( ⇒ x(x+6)-1(x+6) = 0 \)
\( ⇒ (x+6)(x-1) = 0 \)
Either \( x+6 = 0 \)
\( ⇒ x = -6 \)
or \( x-1 = 0 \)
\( ⇒ x = 1 \)
⸫ The required roots are -6 and 1
(vii) \( \sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}} = \frac{13}{6}...(1) \)
Let \( y = \sqrt{\frac{x}{1-x}} ⇒ \frac{1}{y} = \sqrt{\frac{1-x}{x}}\)
⸫ (1) ⇒ \( y+\frac{1}{y} = \frac{13}{6} \)
\( ⇒ \frac{y^2+1}{y}=\frac{13}{6} \)
\( ⇒ 6y^2+6=13y \)
\( ⇒ 6y^2-13y+6 = 0 \)
\( ⇒ 6y^2-(9+4)y+6 = 0 \)
\( ⇒ 6y^2-9y-4y+6 = 0 \)
\( ⇒ 3y(2y-3)-2(2y-3) = 0 \)
\( ⇒ (2y-3)(3y-2) \)
Either \( 2y-3 = 0 ⇒ y = \frac{3}{2} \)
\( ⇒ \sqrt{\frac{x}{1-x}} = \frac{3}{2} \)
\( ⇒ (\sqrt{\frac{x}{1-x}})^2 = (\frac{3}{2})^2 \)
\( ⇒ \frac{x}{1-x}= {9}{4} ⇒ 4x = 9-9x \)
\( ⇒ 13 x = 9 ⇒ x = \frac{9}{13} \)
or \( 3y-3 = 0 ⇒ y = \frac{2}{3} \)
\( ⇒ \sqrt{\frac{x}{1-x}} = \frac{2}{3} \)
\( ⇒ (\sqrt{\frac{x}{1-x}})^2 = (\frac{2}{3})^2 \)
\( ⇒ \frac{x}{1-x}= {4}{9} \)
\( ⇒ 9x = 4-4x \)
\( ⇒ 13x = 4 \)
\( ⇒ x = \frac{4}{13} \)
⸫ The required solutions are \( \frac{9}{13}\) and \(\frac{4}{13} \)
(viii) \( (x^2+\frac{1}{x^2})-5(x+\frac{1}{x}) = 4 \)
\( ⇒ x^2+2(x)(\frac{1}{x})+(\frac{1}{x})^2-5(x+\frac{1}{x}) -2 =4 \)
\( ⇒ (x+\frac{1}{x})^2-5(x+\frac{1}{x})-6 = 0....(1) \)
Let \( x+\frac{1}{x} = y \)
⸫ (1) ⇒ \( y^2-5y-6 = 0 \)
\( ⇒ y^2-(6-1)y-6 = 0 \)
\( ⇒ y^2-6y+y-6 = 0 \)
\( ⇒ y(y-6)+1(y-6) \)
\( ⇒ (y-6)(y+1) \)
Either \( y-6 = 0 \)
\( ⇒ y = 6 \)
\( ⇒ x+\frac{1}{x}= 6 \)
\( ⇒ \frac{x^2+1}{x} = 6 \)
\( ⇒ x^2-6x+1 = 0 \)
⸫ \( x = \frac{-(-6)\pm \sqrt{6^2-4(1)(1)}}{2(1)} \)
\( = \frac{6\pm\sqrt{36-4}}{2} = \frac{6\pm4\sqrt{2}}{2} \)
\( = \frac{2(3\pm2\sqrt{2})}{2} = 3\pm2\sqrt{2} \)
or \( y+1= 0 \)
\( ⇒ x+\frac{1}{x}+1 = 0 \)
\( ⇒ \frac{x^2+1+x}{x} = 0 \)
\( ⇒ x^2+x+1 = 0 \)
⸫ \( x = \frac{-1\pm\sqrt{1^2-4(1)(1)}}{2(1)} \)
\( = \frac{-1\pm\sqrt{-3}}{2} = \frac{-1\pm\sqrt{3}i}{2} \)
⸫ The required solutions are \( 3\pm2\sqrt{2} \) and \( \frac{-1\pm\sqrt{3}i}{2} \)
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(ix) \( (x-1)(x-2)(x-3)(x-4) = 120 \) \( ⇒ [(x-1)(x-4)][(x-2)(x-3)] = 120 \) [⸪ -1-4=-2-3] \( ⇒ (x^2-4x-x+4)(x^2-3x-2x+6) = 120 \) \( ⇒ (x^2-5x+4)(x^2-5x+6) = 120 \) Let \( x^2-5x = y \) ⸫ (1) ⇒ \( (y+4)(y+6) = 120 \) \( ⇒ y^2+6y+4y+36 = 120 \) \( ⇒ y^2+10y-96 = 0 \) \( ⇒ y^2+(16-6)y-96 = 0 \) \( ⇒ Y^2+16y-6y-96 = 0 \) \( ⇒ y(y+16)-6(y+16) = 0 \) \( ⇒ (y+16)(y-6) = 0 \) Either \( y+16 = 0 \) \( ⇒ x^2-5x+16 = 0 \) \( ⇒ x = \frac{-(-5)\pm\sqrt{(-5)^2-4.1.16}}{2.1} \) \( = \frac{5 \pm \sqrt{-39}}{2} = \frac{5\pm\sqrt{39}i}{2} \) or \( y-6 = 0 \) \( ⇒ x^2-5x-6 = 0 \) \( ⇒ x^2-6x+x-6 = 0 \) \( ⇒ x(x-6)+1(x-6) \) \( ⇒ (x+1)(x-6) = 0 \) \( ⇒ x = 6, -1 \) ⸫ The required solutions are 6, \( -1 \) and \( \frac{5\pm\sqrt{39}i}{2} \) |
(x) \( (x+1)(x+3)(x+5)(x+7) = 20 \)
\( ⇒ (x+1)(x+7)(x+3)(x+5) = 20 \)
\( ⇒ (x^2+7x+x+7)(x^2+5x+3x+15) = 20 \)
\( ⇒ (x^2+8x+7)(x^2+8x+15) = 20...(1) \)
Let \( x^2+8x = y \)
⸫ (1) ⇒ \( (y+7)(y+15) = 20 \)
\( ⇒ y^2+15y+7y+105 = 20 \)
\( ⇒ y^2+22y+85 = 0 \)
\( ⇒ y^2+17y+5y+85 = 0 \)
\( ⇒ y(y+17)+5(y+17) = 0 \)
\( ⇒ (y+17)(y+5) = 0 \)
Either \( y+17 = 0 \)
\( ⇒ x^2+8x+17 = 0 \)
\( ⇒ x = \frac{-8\pm\sqrt{8^2-4.17}}{2} \)
\( = \frac{-8\pm\sqrt{64-68}}{2} \)
\( = \frac{-8\pm\sqrt{-4}}{2} \)
\( = \frac{-8\pm2i}{2} \)
\( = \frac{2(-4\pm{i})}{2} \)
\( = -4\pm{i} \)
or \( y+5 = 0 \)
\( ⇒ x^2+8x+5 = 0 \)
\( ⇒ x = \frac{-8\pm\sqrt{8^2-4.5}}{2} \)
\( = \frac{-8\pm\sqrt{64-20}}{2} \)
\( = \frac{-8\pm\sqrt{44}}{2} \)
\( = \frac{-8\pm2\sqrt{11}}{2} \)
\( = \frac{2(-4\pm\sqrt{11})}{2} \)
\( = -4\pm\sqrt{11} \)
⸫ The required solutions are \( -4\pm{i} \) and \( -4\pm\sqrt{11} \)
(xi) \( \frac{1}{x-2}+\frac{1}{x+5}=\frac{1}{x-6} \)
\( ⇒ \frac{x+5+x-2}{(x-2)(x+5)} = \frac{1}{x-6} \)
\( ⇒ \frac{2x+3}{x^2+5x-2x-10} = \frac{1}{x-6} \)
\( ⇒ \frac{2x+3}{x^2+3x-10} = \frac{1}{x-6} \)
\( ⇒ 2x^2-12x+3x-18 = x^2+3x-10 \)
\( ⇒ x^2-12x-8 = 0 \)
\( ⇒ x = \frac{12\pm\sqrt{12^2-4(-8)}}{2} \)
\( = \frac{12\pm\sqrt{144+32}}{2} \)
\( = \frac{12\pm\sqrt{176}}{2} \)
\( = \frac{12\pm4\sqrt{11}}{2} \)
\( = \frac{2(6\pm2\sqrt{11})}{2} \)
\( = 6\pm2\sqrt{11} \)
⸫ The required solutions are \( 6\pm2\sqrt{11} \)
(xii) \( \frac{5}{x^2+6x+8}= \frac{1}{x^2+6x+5}+\frac{4}{x^2+6x+9} \) .....(1)
Let \( x62=6x = y \)
⸫ (1) ⇒ \( \frac{5}{y+8}= \frac{1}{y+5}+\frac{4}{y+9} \)
\( ⇒ \frac{4}{y+8}+\frac{1}{y+8}=\frac{1}{y+5}+\frac{4}{y+9} \)
\( ⇒ \frac{4}{y+8}-\frac{4}{y+9}=\frac{1}{y+5}-\frac{1}{y+8} \)
\( ⇒ 4(\frac{1}{y+8}-\frac{1}{y+9})= \frac{y+8-y-5}{(y+5)(y+8)} \)
\( ⇒ 4(\frac{y+8-y-8}{(y+8)(y+9)}) = \frac{3}{(y+5)(y+8)} \)
\( ⇒ 4(\frac{1}{(y+8)(y+9)})=\frac{3}{(y+5)(y+8)} \)
\( ⇒ 4(y+5)(y+8)=3(y+8)(y+9) \)
\( ⇒ 4(y+5)(y+8)-3(y+8)(y+9) \)
\( ⇒ (y+8)[4(y+5)-3(y+9)]=0 \)
\( ⇒ (y+8)[4y+20-3y-27] = 0 \)
\( ⇒ (y+8)(y-7) = 0 \)
Either \( y+8 = 0 \)
\( ⇒ x^2+6x+8=0 \)
\( ⇒ x^2+2x+4x+8=0 \)
\( ⇒ x(x+2)+4(x+2)=0 \)
\( ⇒ x = -2 \) or \( -4 \)
or \( y-7=0 \)
\( ⇒ x^2+6-7 = 0 \)
\( ⇒ x^2+7x-x-7 = 0 \)
\( ⇒ x(x+7)-1(x+7) =0 \)
\( ⇒ x=1 \)
or \( -7 \)
⸫ The required solutions are \(1, -2, -4 \) and \( -7 \)
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(xiii) \( \frac{x^2-5}{2x-3}-\frac{3x^2}{6x+1} = \frac{3}{2} \) \( ⇒ \frac{x^2-5}{2x-3} = \frac{3}{2} + \frac{3x^2}{6x+1} \) \( ⇒ \frac{x^2-5}{2x-3} = \frac{3(6x+1)+6x^2}{2(6x+1)} \) \( ⇒ \frac{x^2-5}{2x-3} = \frac{18x+3+6x^2}{12x+2} \) \( ⇒ 12x^3+2x^2-60x-10 = 36x^2+6x+12x^3-54x-9-18x^2 \) \( ⇒ 2x^2+18x^2-36x^2-60x+54x-6x+9-10 = 0 \) \( ⇒ -16x^2-12x-1 = 0 \) \( ⇒ 16x^2+12x+1 = 0 \) \( ⇒ x = \frac{-12\pm\sqrt{12^2-4.16}}{2.16} \) \( = \frac{-12\pm\sqrt{144-64}}{32} \) \( = \frac{-12\pm\sqrt{80}}{32} \) \( = \frac{-12\pm4\sqrt{5}}{32} \) \( = \frac{4(-3\pm\sqrt{5})}{32} \) \( = \frac{-3\pm\sqrt{5}}{8} \) ⸫ The required solutions are \( \frac{-3\pm\sqrt{5}}{8} \) |
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(xiv) \( \sqrt{2x-1}+\sqrt{3x-2}=\sqrt{4x-3}+\sqrt{5x-4} \) \( ⇒ \sqrt{2x-1}-\sqrt{5x-4}=\sqrt{4x-3}-\sqrt{3x-2} \) \( ⇒ (\sqrt{2x-1}-\sqrt{5x-4})^2=(\sqrt{4x-3}-\sqrt{3x-2})^2 =4x-3+3x-2-2(\sqrt{4x-3})(\sqrt{3x-2}) \) \( ⇒ 7x-5-2(\sqrt{2x-1})(\sqrt{5x-4})= 7x-5-2(\sqrt{4x-3})(\sqrt{3x-2}) \) \( ⇒ \sqrt{2x-1}.\sqrt{5x-4}=\sqrt{4x-3}.\sqrt{3x-2} \) \( ⇒ (\sqrt{2x-1}.\sqrt{5x-4})^2=(\sqrt{4x-3}.\sqrt{3x-2})^2 \) \( ⇒ (2x-1)(5x-4)=(4x-3)(3x-2) \) \( ⇒ 10x^2-8x-5x+4=12x^2-8x-9x+6 \) \( ⇒ -2x^2+4x-2=0 \) \( ⇒ x^2-2x+1=0 \) \( ⇒ (x-1)^2 =0 \) \( ⇒ x=1, 1 \) ⸫ The required solutions are \( 1\) and \( 1\) |
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(xv) \( \sqrt{x^2-4}+\sqrt{x^2+5x+6}=\sqrt{3x^2+13x+14} \) \( ⇒ \sqrt{(x+2)(x-2)}+\sqrt{(x+2){x+3}}-{\sqrt{x+2}(3x+7)} = 0 \) \( ⇒ \sqrt{x+2}(\sqrt{x-2}+\sqrt{x+3}-\sqrt{3x+7}) = 0 \) Either \( \sqrt{x+2} = 0 \) \( ⇒ x+2 = 0 \) \( ⇒ x = -2 \) or \( \sqrt{x-2}+\sqrt{x+3}-\sqrt{3x+7} = 0 \) \( ⇒ \sqrt{x-2}+\sqrt{x+3}=\sqrt{3x+7} \) \( ⇒ (\sqrt{x-2}+\sqrt{x+3})^2=(\sqrt{3x+7})^2 \) \( ⇒ (\sqrt{x-2})^2+(\sqrt{x+3})^2+2.\sqrt{x-2}.\sqrt{x+3}=3x+7 \) \( ⇒ x-2+x+3+2.\sqrt{(x-2)(x+3)}=3x+7 \) \( ⇒ 2.\sqrt{(x-2)(x+3)} = x+6 \) \( ⇒ (2.\sqrt{(x^2+x-6})^2 = (x+6)^2 \) \( ⇒ 4(x^2+x-6) = x^2+36+12x \) \( ⇒ 4x^2+4x-24-x^2-12x-36=0 \) \( ⇒ 3x^2-8x-60=0 \) \( ⇒ 3x^2-18x+10x-60=0 \) \( ⇒ 3x(x-6)+10(x-6) = 0 \) \( ⇒ (x-6)(x+10) = 0 \) Either \( x = 6 \) or \( x = -\frac{10}{3} \) [Unacceptable] ⸫ The required solutions are \(-2\) and \(6\) |
(xvi) \( 3^{x+3}+3^x-3^{2x+1} = 9 \)
\( ⇒ 3^x.3^3+3^x-3^x.3 =9 \)
\( ⇒ 27.3^x+3^x-3.3^{2x} = 9....(1) \)
Let \( 3^x = y \)
⸫ (1) ⇒ \( 27y+y-3y^2 = 9 \)
\( ⇒ -3y^2+28y-9 = 0 \)
\( ⇒ 3y^2-28y+9 = 0 \)
\( ⇒ 3y^2-(1+27)y+9 = 0 \)
\( ⇒ 3y^2-y-27y+9 = 0 \)
\( ⇒ y(3y-1)-9(3y-1) \)
\( ⇒ (y-9)(3y-1) \)
Either \( y-9 = 0 \)
\( ⇒ 3^x=9=3^2 \)
\( ⇒ x = 2 \)
or \( 3y-1 = 0 \)
\( ⇒ y =\frac{1}{3} \)
\( ⇒ 3^x = 3^{-1} \)
\(⇒ x = -1 \)
⸫ The required solutions are \(2\) and \(-1\)
(xvii) \( 4^x-3.2^{x+2}+32 = 0 \)
\( ⇒ (2)^{2x}-3.2^x.2^2+32 = 0 \)
\( ⇒ 2^{2x}+12.2^{2x}+32 = 0...(1) \)
Let \( 2^x = y \)
⸫ (1) ⇒ \( y^2-12y+32 = 0 \)
\( ⇒ y^2-4y-8y+32 = 0 \)
\( ⇒ y(y-4)-8(y-4) = 0 \)
\( ⇒ (y-4)(y-8) = 0 \)
Either \( y-4 = 0 \)
\( ⇒ y = 4 \)
\( ⇒ 2^x = 4 \)
\( ⇒ 2^x = 2^2 \)
\( ⇒ x = 2 \)
or \( y-8 = 0 \)
\( ⇒ y = 8 \)
\( ⇒ 2^x = 2^3 \)
\( ⇒ x = 3 \)
⸫ The required solutions are 2 and 3
(xviii) \( x^{\frac{2}{3}}-x^{\frac{1}{3}}-2 = 0...(1) \)
Let \( x^{\frac{1}{3}} = y \)
⸫ (1) ⇒ \( y^2-y-2 = 0 \)
\( ⇒ y^2-(2-y)y-2 = 0 \)
\( ⇒ y^2 -2y+y-2 = 0 \)
\( ⇒ y(y-2)+1(y-2) = 0 \)
\( ⇒ (y-2)(y+1) = 0 \)
Either \( y-2 = 0 \)
\( ⇒ x^{\frac{1}{3}} = 2 \)
\( ⇒ (x^{\frac{1}{3}})^3 = 2^3 \)
\( ⇒ x = 8 \)
or \( y+1 = 0 \)
\( ⇒ x^{\frac{1}{3}} = -1 \)
\( ⇒ (x^{\frac{1}{3}})^3 = (-1)^3 \)
\( ⇒ x = -1 \)
⸫ The required solutions are \(-1\) and \(8\)
(xix) \( x^{-4}-10x^{-2}+9 = 0...(1) \)
Let \( x^{-2} = y \)
⸫ (1) ⇒ \( y^2-10y+9 = 0 \)
\( ⇒ y^2-9y-y+9 = 0 \)
\( ⇒ y(y-9)-1(y-9) = 0 \)
\( ⇒ (y-9)(y-1) = 0 \)
Either \( y-9 = 0 \)
\( ⇒ x^{-2} = 9 \)
\( ⇒ x^2 = \frac{1}{9} \)
\( ⇒ x = \pm \frac{1}{3} \)
or \( y-1 = 0 \)
\( ⇒ x^{-2} = 1 \)
\( ⇒ x^2 = 1 \)
\( ⇒ x = \pm1 \)
⸫ The required solutions are \(1, -1, \frac{1}{3} \) and \(-\frac{1}{3} \)
(xx) \( 3^{2x}+9 = 10(\frac{1}{3})^{-x} \)
\( ⇒ 3^{2x}+9 = 10.3^x \)
\( ⇒ 3^{2x}-10.3^x+9 = 0...(1) \)
Let \( 3^x = y \)
⸫ (1) ⇒ \( y^2-10y+9 = 0 \)
\( ⇒ y^2-9y-y+9 = 0 \)
\( ⇒ y(y-9)-1(y-9) = 0 \)
\( ⇒ (y-9)(y-1) = 0 \)
Either \( y-9 = 0 \)
\( ⇒ 3^x = 9 \)
\( ⇒ 3^x = 3^2 \)
\( ⇒ x = 2 \)
or \( y-1 = 0 \)
\( ⇒ 3^x = 1 \)
\( ⇒ 3^x = 3^0 \)
\( ⇒ x = 0 \)
⸫ The required solutions are 0 and 2