Exercise 4.2 | Class 10 | Advanced Mathematics | Quadratic Equation

1. Which number when added to its square will make 156?
Solution:
Let the number be $x$
According to question,
$x+x^2=156$
⇒ $x^2+x-156=0$
⇒ $x^2+13x-12x-156=0$
⇒ $x(x+13)-12(x+13)=0$
⇒ $(x+13)(x-12)=0$
⇒ $x=-13$ [Rejected] or $x=12$
⸫ The required number is $12$.



2. If the sum of the squares of two consecutive odd natural numbers is $290$. What are the numbers?
Solution:
Let the numbers be $x$ and $x+2$
According to question,
$x^2+(x+2)^2=290$
⇒ $x^2+x^2+4+4x-290=0$
⇒ $2x^2+4x-286=0$
⇒ $x^2+2x-143=0$   [Dividing the equation by $2$]
⇒ $x^2+13x-11x-143=0$
⇒ $x(x+13)-11(x+13)=0$
⇒ $(x+13)(x-11)=0$
⇒ $x=-13$ [Rejected] or $x=11$
⸫ The required numbers are $11$ and $13$.



3. Divide the number $18$ into two positive parts so that the sum of their squares is equal to $15$ times of the larger part.
Solution:
Let the larger of the two parts be $x$ and the smaller part be $18-x$
According to question,
$x^2+(18-x)^2=15x$
⇒ $x^2+324-36x+x^2-15x=0$
⇒ $2x^2-51x-324=0$
⇒ $2x^2-27x-24x+324=0$
⇒ $x(2x-27)-12(2x-27)=0$
⇒ $(2x-27)(x-12)=0$
⇒ $x=\frac{27}{2}$ [Rejected] or $x=12$
⸫ The required larger part of the number is $12$ and the smaller part is $18-12=6$.



4. The product of the digits of a two digit number is 12. If 36 is added to the number, the digits interchange their position. What is the number?
Solution:
Let the tens digit of the number be $x$ and the ones digit be $y$
⸫ The number = $10x+y$
Also, $xy=12 ⇒ y= \frac{12}{x}$ ...(1)
According to question,
$(10x+y)+36=10y+x$
⇒ $10x+y+36=10y+x$
⇒ $9x-9y=-36$
⇒ $x-y=-4$  [Dividing the equation by $9$]
⇒ $x-\frac{12}{x}=-4$  [using (1)]
⇒ $\frac{x^2-12}{x}=-4$
⇒ $x^2-12=-4x$
⇒ $x^2+4x-12=0$
⇒ $x^2+6x-2x-12=0$
⇒ $x(x+6)-2(x+6)=0$
⇒ $(x+6)(x-2)=0$
⇒ $x=-6$ [Rejected] or $x=2$
Now, (1)⇒ $y=\frac{12}{2}=6$
⸫ The required number is $(10×2)+6=26$.


5. The sum of the two numbers is $25$ and the sum of their cubes is $4825$. What are the numbers?
Solution:
Let the numbers be $x$ and $25-x$
According to question,
$x^3+(25-x)^3=4825$
⇒ $(x+25-x)[x^x(25-x)+(25-x)^2]=4825$
⇒ $x^2-25x+x^2+625-50x+x^2=4825$
⇒ $3x^2-75x=193-625$
⇒ $3x^2-75x+432=0$
⇒ $x^2-25x+144=0$
⇒ $x^2-16x-9x+144=0$
⇒ $x(x-16)-9(x-16)=0$
⇒ $(x-16)(x-9)=0$
⇒ $x=16$ or $x=9$
⸫ The required numbers are $16$ and $9$.



6. The perimeter of a rectangular field is $88~m$ and its area is $420~m^2$. Find the length and breadth of the rectangular field.
Solution:
Here, Perimeter of the rectangular field = $88~m$
⇒ $2(length+breadth)=88$
⇒ $length+breadth=44$
Let us now consider the length of the rectangular field be $x~m$ and so the breadth be $(44-x)~m$
According to question,
Area of the rectangular field = $420~m^2$
⇒ $x(44-x)=420$
⇒ $44x-x^2=420$
⇒ $x^2-44x+420=0$
⇒ $x^2-30x-14x+420=0$
⇒ $x(x-30)-14(x-30)=0$
⇒ $(x-30)(x-14)=0$
⇒ $x=30$ or $x=14$
When $x=30$,
The length of the rectangular field is $30~m$ and breadth is $44-30=14~m$.
When $x=14$,
The length of the rectangular field is $14~m$ and breadth is $44-14=30~m$.



7. The sum of the areas of two squares is $225~cm^2$. The length of the side of one is $3~cm$ more than the other. Find the sides of two squares.
Solution:
Let the side of one of the squares be $x~cm$ and that of the other be $(x+3)~cm$
According to question,
$x^2+(x+3)^2=225$
⇒ $x^2+x^2+9+6x-225=0$
⇒ $2x^2+6x-216=0$
⇒ $x^2+3x-108=0$   [Dividing the equation by $2$]
⇒ $x^2+12x-9x-108=0$
⇒ $x(x+12)-9(x+12)=0$
⇒ $(x+12)(x-9)=0$
⇒ $x=-12$ [Rejected] or $x=9$
⸫ The required side of one of the squares is $9~cm$ and that of the other be $(9+3)=12~cm$



8. The price of a long piece of cloth is Rs. $440$. If the cloth would have been $1$ metre longer and the price per metre would have been Rs. $4$ less, then cost of the cloth does not alter. What was the length of the piece of the cloth?
Solution:
Let the length of the cloth be $x~m$
Given, total price of the cloth= Rs. $440$
Cost of each metre = Rs. $\frac{440}{x}$
When the length is extended by $1~m$,
length of the cloth becomes $(x+1)~m$
Cost of each metre in such case = Rs. $\frac{440}{x+1}$
According to question,
$\frac{440}{x+1}=\frac{440}{x}-4$
⇒ $4=\frac{440}{x}-\frac{440}{x+1}$
⇒ $4=\frac{440(x+1)-440x}{x(x+1)}$
⇒ $4x(x+1)=440x+440-440x$
⇒ $4x^2+4x-440=0$
⇒ $x^2+x-110=0$
⇒ $x^2+11x-10x-110=0$
⇒ $x(x+11)-10(x+11)=0$
⇒ $(x+11)(x-10)=0$
⇒ $x=-11$ [Rejected] or $x=10$
⸫ The required length of the cloth is $10~m$

Alternative Solution:
Let the length of the cloth be $x~m$
According to question,
$(x+1)\left(\frac{440}{x}-4\right)=440$
⇒ $(x+1)\left(\frac{440-4x}{x}\right)=440$
⇒ $(x+1)(440-4x)=440x$
⇒ $440x-4x^2+440-4x=440x$
⇒ $-4x^2+440-4x=0$
⇒ $4x^2+4x-440=0$
⇒ $x^2+x-110=0$
⇒ $x^2+11x-10x-110=0$
⇒ $x(x+11)-10(x+11)=0$
⇒ $(x+11)(x-10)=0$
⇒ $x=-11$ [Rejected] or $x=10$
⸫ The required length of the cloth is $10~m$



9. Rs. $1500$ is distributed among some labours. If the number of labours would have been $5$ more, then each of them would have been received Rs. $15$ less. Find the number of labours.
Solution:
Let the number of labour be $x$
Given, total amount distributed = Rs. $1500$
Amount received by each labour = Rs. $\frac{1500}{x}$
When the number of labour is increased by $5$,
Number of labour becomes $x+5$
Amount received by each labour in such case = Rs. $\frac{1500}{x+5}$
According to question,
$\frac{1500}{x+5}=\frac{1500}{x}-15$
⇒ $15=\frac{1500}{x}-\frac{1500}{x+5}$
⇒ $15=\frac{1500(x+5)-1500x}{x(x+5)}$
⇒ $15x(x+5)=1500x+7500-1500x$
⇒ $15x^2+75x-7500=0$
⇒ $x^2+5x-500=0$
⇒ $x^2+25x-20x-500=0$
⇒ $x(x+25)-20(x+25)=0$
⇒ $(x+25)(x-20)=0$
⇒ $x=-25$ [Rejected] or $x=20$
⸫ The required number of labour is $20$

Alternative Solution:
Let the number of labour be $x$
According to question,
$(x+5)\left(\frac{1500}{x}-15\right)=1500$
⇒ $(x+5)\left(\frac{1500-15x}{x}\right)=1500$
⇒ $(x+5)(1500-15x)=1500x$
⇒ $1500x-15x^2+7500-75x=1500x$
⇒ $-15x^2+7500-75x=0$
⇒ $15x^2+75x-7500=0$
⇒ $x^2+5x-500=0$
⇒ $x^2+25x-20x-500=0$
⇒ $x(x+25)-20(x+25)=0$
⇒ $(x+25)(x-20)=0$
⇒ $x=-25$ [Rejected] or $x=20$
⸫ The required number of labour is $20$



10. Find the value:
(i) $\sqrt{20+\sqrt{20+\sqrt{20+....... \infty}}}$
Solution:
Let $x=\sqrt{20+\sqrt{20+\sqrt{20+....... \infty}}}$
⇒ $x^2=20+\sqrt{20+\sqrt{20+..... \infty}}$
⇒ $x^2-20-x=0$
⇒ $x^2-x-20=0$
⇒ $x^2-5x+4x-20=0$
⇒ $x(x-5)+4(x-5)=0$
⇒ $(x-5)(x+4)=0$
⇒ $x=5$ or $x=-4$


(ii) $\sqrt{6+\sqrt{6+\sqrt{6+....... \infty}}}$
Solution:
Let $x=\sqrt{6+\sqrt{6+\sqrt{6+....... \infty}}}$
⇒ $x^2=6+\sqrt{6+\sqrt{6+..... \infty}}$
⇒ $x^2-6-x=0$
⇒ $x^2-x-6=0$
⇒ $x^2-3x+2x-6=0$
⇒ $x(x-3)+2(x-3)=0$
⇒ $(x-3)(x+2)=0$
⇒ $x=3$ or $x=-2$


(iii) $\sqrt{3+\sqrt{3+\sqrt{3+....... \infty}}}$
Solution:
Let $x=\sqrt{3+\sqrt{3+\sqrt{3+....... \infty}}}$
⇒ $x^2=3+\sqrt{3+\sqrt{3+..... \infty}}$
⇒ $x^2-3-x=0$
⇒ $x^2-x-3=0$
Using quadratic formula, we have
$x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(3)}}{2×1}$
$x=\frac{1\pm\sqrt{1+12}}{2}$
$x=\frac{1\pm\sqrt{13}}{2}$
Either $x=\frac{1+\sqrt{13}}{2}$ or $x=\frac{1-\sqrt{13}}{2}$


(iv) $2+\frac{1}{2+\frac{1}{2+...... \infty}}$
Solution:
Let $x=2+\frac{1}{2+\frac{1}{2-...... \infty}}$
⇒ $x^2=2+\frac{1}{x}$
⇒ $x^2-2x-1=0$
Using quadratic formula, we have
$x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-1)}}{2×1}$
$x=\frac{2\pm\sqrt{4+4}}{2}$
$x=\frac{2\pm\sqrt{8}}{2}$
$x=\frac{2\pm2\sqrt{2}}{2}$
$x=\frac{2(1\pm\sqrt{2})}{2}$
⇒ $x=1\pm\sqrt{2}$
Either $x=1+\sqrt{2}$ or $x=1-\sqrt{2}$


(v) $2-\frac{1}{2-\frac{1}{2-...... \infty}}$
Solution:
Let $x=2-\frac{1}{2-\frac{1}{2-...... \infty}}$
⇒ $x^2=2-\frac{1}{x}$
⇒ $x^2-2x+1=0$
⇒ $(x-1)^2=0$
⇒ $x=1$

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