Exercise 4.1 | Class 10 | Advanced Mathematics | Quadratic Equation

1. Form a quadratic equation using the following roots:

(i) $-5, 7$ Solution: Here, the given roots are $5$ and $7$ ⸫ The required quadratic equation is, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $\ ⇒ x^2-[(-5)+7]x+(-5)7=0$ $\ ⇒ x^2-2x-35=0$

(ii) $-\frac{1}{2}, -3$ Solution: Here, the given roots are $-\frac{1}{2}$ and $-3$ ⸫ The required quadratic equation is, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $\ ⇒ x^2 -[(-\frac{1}{2})+(-3)]x+(-\frac{1}{2})(-3)=0$ $\ ⇒ x^2-(-\frac{7}{2})x +\frac{3}{2}=0$ $\ ⇒ x^2+\frac{7}{2} x+\frac{3}{2}=0$ $\ ⇒ 2x^2+7x+3=0$

(iii) $1, -\frac{3}{2}$ Solution: Here, the given roots are $1$ and $-\frac{3}{2}$ ⸫ The required quadratic equation is, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $\ ⇒ x^2-[(1)+(-\frac{3}{2})]x+(1)(-\frac{3}{2})=0$ $\ ⇒ x^2-(-\frac{1}{2})x-\frac{3}{2}=0$ $\ ⇒ x^2+\frac{1}{2}x-\frac{3}{2}=0$ $\ ⇒ 2x^2+x-3=0$

(iv) $1, -\frac{4}{5}$ Solution: Here, the given roots are $1$ and $-\frac{4}{5}$ ⸫ The required quadratic equation is, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $\ ⇒ x^2-[(1)+(-\frac{4}{5})]x+(1)(-\frac{4}{5})=0$ $\ ⇒ x^2-\frac{1}{5} x-\frac{4}{5}=0$ $\ ⇒ x^2-\frac{1}{5}x-\frac{4}{5}=0$ $\ ⇒ 5x^2-x-4=0$

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $\ ⇒ x^2-[(\frac{1}{2})+(-\frac{1}{3})]x+(\frac{1}{2})(-\frac{1}{3})=0$ $\ ⇒ x^2-\frac{1}{6} x-\frac{1}{6}=0$ $\ ⇒ x^2-\frac{1}{6} x-\frac{1}{6}=0$ $\ ⇒ 6x^2-x-1=0$

(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

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $⇒ x^2-[(\sqrt{3}i)+(-\sqrt{3}i)]x+(\sqrt{3}i)(-\sqrt{3}i) = 0$ $⇒ x^2-[0]x + [\sqrt{9} i^2] =0$ $⇒ x^2-0 + [-3(-1)]=0$ $⇒ x^2+3=0$

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $⇒ x^2- [(4+\sqrt{5})+(4-\sqrt{5})]x+[(4+\sqrt{5})(4-\sqrt{5})]=0$ $⇒ x^2-[8]x+[(4)^2-(\sqrt{5})^2]=0$ $⇒ x^2-8x+[16-5]=0$ $⇒ x^2-8x+11=0$

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $⇒ x^2-[(2-\sqrt{3})+(2+\sqrt{3})]x+[(2-\sqrt{3})(2+\sqrt{3})]=0$ $⇒ x^2-[4]x+[(2)^2-(\sqrt{3})^2]=0$ $⇒ x^2-4x+[4-3]=0$ $⇒ x^2-4x+1=0$

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $⇒ x^2-[(1-\sqrt{\frac{3}{2}}i)+(1+\sqrt{\frac{3}{2}}i)]x+[(1-\sqrt{\frac{3}{2}}i)(1+\sqrt{\frac{3}{2}}i)]=0$ $⇒ x^2-[2]x+[(1)^2-(\sqrt{\frac{3}{2}} i)^2]=0$ $⇒ x^2-2x+[1-[\frac{3}{4 (-1)}] ]=0$ $⇒ x^2-2x+[1+\frac{3}{4}]=0$ $⇒ x^2-2x+\frac{7}{4}=0$ $⇒ 4x^2-8x+7=0$

(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, $x^2-(sum~of~the~roots)x+(product~of~the~roots)=0$ $⇒ x^2-[(\frac{-1-\sqrt{5}}{2})+(\frac{-1+\sqrt{5}}{2})]x+[(\frac{-1-\sqrt{5}}{2})(\frac{-1+\sqrt{5}}{2})]=0$ $⇒ x^2-[\frac{-1-\sqrt{5}-1+\sqrt{5}}{2}] x+[\frac{(-1)^2-(\sqrt{5})^2}{4}]=0$ $⇒ x^2-[\frac{-2}{2}] x+[\frac{1-5}{4}]=0$ $⇒ x^2-x + [\frac{-4}{4}]=0$ $⇒ x^2-x-1 =0$

(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

(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.)

(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.)

(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

(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)] $⇒ 2×\frac{25}{36} = \frac{k}{2}$ $⇒ \frac{25}{18} = \frac{k}{2}$ $⇒ k = \frac{25}{9}$

(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

(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 $m = \frac{7}{4}, m = \frac{-11}{8}$

(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)] ⇒ $\frac{b^2}{a^2(1+n)^2}=\frac{c}{na}$ ⇒ $\frac{b^2}{ac}=\frac{(1+n)^2}{n}$

(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)] ⇒ $\frac{p^2}{9}=\frac{q}{2}$ ⇒ $\frac{p^2}{q}=\frac{9}{2}$

(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)] ⇒ $mn\left[\frac{b^2}{a^2(m+n)^2}\right]=\frac{c}{a}$ ⇒ $mnb^2=ac(m+n)^2$

(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.

(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)] ⇒ $4\left[\frac{b^2}{25a^2}\right]=\frac{c}{a}$ ⇒ $4b^2=25ac$

(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 $\frac{-1\pm\sqrt{3}i}{2}$

(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 $-7$

(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 $\frac{-3\pm\sqrt{5}}{8}$

(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 $1$

(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

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