RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.3

RD Sharma Class 10 Solutions Chapter 8 Ex 8.3 PDF Free Download

Exercise 8.3

 

Question 1: Find the roots of the equation (x – 4) (x + 2) = 0

Solution:

We have given with the equation : (x – 4) (x + 2) = 0

Either x – 4 = 0, x = 4

Or, x + 2 = 0, x = -2

Therefore, the roots of the given quadratic equation are  4 and -2.

 

Question 2: Find the roots of the equation (2x + 3) (3x – 7) = 0

Solution:

We have given with the equation : (2x + 3) (3x – 7) = 0.

Either 2x+3 =0, so x = -3/2

Or, 3x-7 =0, so x = 7/3

Therefore, the roots of the given quadratic equation are x=-3/2and x= 7/3, respectively.

 

Question 3: Find the roots of the quadratic equation 3x2-14x-5 = 0.

Solution:

We have given with the equation : 3x2-14x-5 = 0

= 3x2-14x-5 = 0

= 3x2-15x+x-5 = 0

= 3x(x-5)+1(x-5) =0

= (3x+1)(x-5) = 0

Either 3x+1 =0; x = -1/3

Or, x-5 =0; x=5

Therefore, the roots of the given quadratic equation are 5 and \(x=\frac{-1}{3}\) respectively.

 

Question 4: Find the roots of the equation 9x2-3x-2=0.

Solution:

We have given with the equation :9x2-3x-2 =0.

= 9x2-3x-2 =0.

= 9x2 -6x+3x-2 =0

= 3x (3x-2)+1(3x-2) =0

= (3x-2)(3x+1) = 0

Either, 3x-2 =0; x=2/3

Or, 3x+1 = 0; x=-1/3

Therefore, the roots of the given quadratic equation are x=2/3 and x=-1/3, respectively.

 

Question 5: Find the roots of the quadratic equation \(\frac{1}{x-1}-\frac{1}{x+5}= \frac{6}{7}\)

Solution:

We have given with the equation :\(\frac{1}{x-1}-\frac{1}{x+5}= \frac{6}{7}\)

= \(\frac{1}{x-1}-\frac{1}{x+5}= \frac{6}{7}\)

= \(\frac{x+5-x+1}{(x-1)(x+5)}= \frac{6}{7}\)

= \(\frac{6}{x^{2}+4x-5}= \frac{6}{7}\)

Cancel the like terms of the numerator, present on both the sides. We get;

= \(\frac{1}{x^{2}+4x-5}= \frac{1}{7}\)

= x2+4x-5 = 7

= x2+4x-12 = 0

= x2+6x-2x-12 = 0

= x(x+6)-2(x-6) =0

=(x+6)(x-2) =0

Either x+6 =0; x= -6

Or, x-2 =0; x=2

Therefore, the roots of the given quadratic equation are 2 and -6 respectively.

 

Question 6: Find the roots of the equation 6x2+11x+3=0.

Solution:

We have given with the equation : 6x2+11x+3 =0.

= 6x2+11x+3 =0.

= 6x2 +9x+2x+3 =0

= 3x(2x+3) +1(2x+3) =0

= (2x+3)(3x+1) =0

Either, 2x+3 =0 ; x = -3/2

Or, 3x+1 =0 ; x = 1/3

Therefore, the roots of the given quadratic equation are x = -3/2 and x = 1/3 respectively .

 

Question 7: Find the roots of the equation 5x2-3x-2=0.

Solution:

We have given with the equation : 5x2-3x-2=0.

= 5x2-3x-2=0.

= 5x2 -5x+2x-2 = 0

= 5x(x-1) +2(x-1) =0

= (5x+2) (x-1) =0

Either 5x+2 =0; x = -2/5

Or, x-1 =0; x = 1

Therefore, the roots of the given quadratic equation are x=1 and x = -2/5, respectively.

 

Question 8: Find the roots of the equation 48x2-13x-1=0.

Solution:

We have given with the equation : 48x2-13x-1=0.

= 48x2-13x-1=0.

= 48x2-16x+3x-1=0.

= 16x (3x-1) +1(3x-1) =0

= (16x+1)(3x-1) =0

Either 16x+1 =0; x = -1/16

Or, 3x-1 =0; x = 1/3

Therefore, the roots of the given quadratic equation are x = -1/16 and x = 1/3, respectively.

 

Question 9: Find the roots of the equation 3x2=-11x-10.

Solution:

We have given with the equation : 3x2=-11x-10

= 3x2=-11x-10

= 3x2+11x+10 = 0

= 3x2+6x+5x+10 =0

= 3x(x+2) +5(x+2) =0

= (3x+2)(x+2) =0

Either 3x+2 =0; x = -2/3

Or, x+2 =0; x= -2

Therefore, the roots of the given quadratic equation are x = -2/3 and x= -2, respectively.

 

Question 10: Find the roots of the equation 25x(x+1) =-4.

Solution:

We have given with the equation : 25x(x+1) =4

= 25x(x+1) =-4

= 25x2+25x +4 = 0

= 25x2+20x+5x+4 =0

= 5x (5x+4)+1(5x+4)=0

= (5x+4)(5x+1) =0

Either 5x+4 = 0; x=-4/5

Or, 5x+1 =0 ; x = -1/5

Therefore, the roots of the given quadratic equation are x=-4/5 and x = -1/5, respectively.

 

Question 12: Find the roots of the quadratic equation \(\frac{1}{x}-\frac{1}{x-2}=3\).

Solution:

We have given with the equation :\(\frac{1}{x}-\frac{1}{x-2}=3\)

=> \(\frac{1}{x}-\frac{1}{x-2}=3\)

=> \(\frac{x-2-x}{x(x-2)}=3\)

=> \(\frac{2}{x(x-2)}=3\)

On cross multiplication, we get;

=>2 = 3x(x-2)

=> 2 = 3x2-6x

=> 3x2-6x-2 = 0

=> 3x2-3x-3x-2 = 0

=> \(3x^{2}-(3+\sqrt{3})x-(3-\sqrt{3})x+[(\sqrt{3}^{2})-1^{2}] = 0 \)

=> \(3x^{2}-(3+\sqrt{3})x-(3-\sqrt{3})x+[(\sqrt{3}^{2})-1^{2}][(\sqrt{3}^{2})-1^{2}] = 0 \)

=> \(\sqrt{3}^{2}x^{2}-\sqrt{3}(\sqrt{3}+1)x-\sqrt{3}(\sqrt{3}-1)x+(\sqrt{3}+1)(\sqrt{3}-1) = 0 \)

=> \(\sqrt{3}x(\sqrt{3}+1)x-(\sqrt{3}x-(\sqrt{3}+1))(\sqrt{3}-1) = 0\)

=> \((\sqrt{3}x-\sqrt{3}-1)(\sqrt{3}x-\sqrt{3}+1)(\sqrt{3}-1)=0\)

Either, \((\sqrt{3}x-\sqrt{3}-1) \); \(x=\frac{\sqrt{3}+1}{\sqrt{3}}\)

Or, \( (\sqrt{3}x-\sqrt{3}+1)(\sqrt{3}-1)\); \(x=\frac{\sqrt{3}-1}{\sqrt{3}}\)

Therefore, the roots of the given quadratic equation are\(x=\frac{\sqrt{3}-1}{\sqrt{3}}\) and \(x=\frac{\sqrt{3}+1}{\sqrt{3}}\) respectively.

 

Question 13: Find the roots of the quadratic equation \(x-\frac{1}{x}=3\).

Solution:

We have given with the equation :\(x-\frac{1}{x}=3\)

= \(x-\frac{1}{x}=3\)

= \(\frac{x^{2}-1}{x}=3\)

= x2-1 = 3x

= x2-1 -3x=0

= \(x^{2}-(\frac{3}{2}+\frac{3}{2})x-1=0\)

= \(x^{2}-\frac{3+\sqrt{3}}{2}x-\frac{3-\sqrt{3}}{2}x-1=0\)

= \(x^{2}-\frac{3+\sqrt{3}}{2}x-\frac{3-\sqrt{3}}{2}x-\frac{-4}{4}=0\)

= \(x^{2}-\frac{3+\sqrt{3}}{2}x-\frac{3-\sqrt{3}}{2}x-\frac{9-13}{4}=0\)

= \(x^{2}-\frac{3+\sqrt{3}}{2}x-\frac{3-\sqrt{3}}{2}x-\frac{(3)^{2}-(\sqrt{13}^{2})}{(2)^{2}}=0\)

= \(x^{2}-\frac{3+\sqrt{3}}{2}x-\frac{3-\sqrt{3}}{2}x+(\frac{3+\sqrt{13}}{2})(\frac{3-\sqrt{13}}{2})=0\)

= \((x-\frac{3+\sqrt{13}}{2})(x-\frac{3-\sqrt{13}}{2})=0\)

Either, \((x-\frac{3+\sqrt{13}}{2})=0\) ; \(\frac{3+\sqrt{13}}{2}\)

Or, \((x-\frac{3-\sqrt{13}}{2})=0\); \(\frac{3-\sqrt{13}}{2}\)

Therefore, the roots of the given quadratic equation are\(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\) respectively.

 

Question 14: Find the roots of the quadratic equation \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\).

Solution:

We have given with the equation :\(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\)

=> \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\)

=> \(\frac{x-7-x-4}{(x+4)(x-7)}=\frac{11}{30}\)

=> \(\frac{-11}{(x+4)(x-7)}=\frac{11}{30}\)

Cancel the like numbers on both the sides of the equation;

=> \(\frac{-1}{(x+4)(x-7)}=\frac{1}{30}\)

=> x2-3x-28 =-30

=> x2-3x-2 =0

=> x2-2x-x-2 =0

=> x(x-2)-1(x-2) =0

=> (x-2)(x-1) =0

Either x-2 = 0; x= 2

Or, x-1 = 0; x= 1

Therefore, the roots of the given quadratic equation are x=1 and x=2, respectively.

 

Question 15: Find the roots of the quadratic equation a2x2-3abx+2b2=0

Solution:

We have given with the equation : a2x2-3abx+2b2=0

= a2x2-3abx+2b2=0

= a2x2-abx-2abx+2b2=0

= ax(ax-b)-2b(ax-b) =0

= (ax-b)(ax-2b) =0

Either ax-b=0; x=b/a

Or, ax-2b =0; x=2b/a

Therefore, the roots of the given quadratic equation are x=b/a and  x=2b/a, respectively.

 

Question 16: Find the roots of the 4x2+4bx-(a2-b2) =0.

Solution:

-4(a2-b2) = -4(a-b)(a+b)

=> -2(a-b) * 2(a+b)

=> 2(b-a) * 2(b+a)

=> 4x2+ (2(b-a) + 2(b+a)) – (a-b)(a+b) = 0

=> 4x+ 2(b-a)x++ 2(b+a)x+(b-a)(a+b) =0

=> 2x(2x+(b-a)) +(a+b)(2x+(b-a)) = 0

=> (2x+(b-a))(2x+b+a) = 0

Either, (2x+(b-a)) = 0; x = (a-b)/2

Or, (2x+b+a) =0; x=-a-b/2

Therefore, the roots of the given quadratic equation are x = (a-b)/2 and x=-(a+b)/2, respectively.

 

Question 17: Find the roots of the equation ax2+(4a2-3b)x -12ab =0.

Solution:

We have given with the equation : ax2+(4a2-3b)x -12ab =0

= ax2+(4a2-3b)x -12ab =0

= ax2+4a2x-3bx -12ab =0

= ax(x-4a) – 3b(x-4a) =0

= (x-4a)(ax-4b) = 0

Either x-4a =0; x = 4a

Or, ax-4b = 0; x=4b/a

Therefore, the roots of the given quadratic equation are x=4b/aand 4a respectively.

 

Question 18: Find the roots of \(\frac{x+3}{x+2}=\frac{3x-7}{2x-3}\).

Solution:

We have given with the equation :\(\frac{x+3}{x+2}=\frac{3x-7}{2x-3}\)

= =(x+3)(2x-3)=(x+2){3x-7)

=2x2-3x+6x-9=3x2-x-14

=2x2+3x-9=3x2-x-14

=x2-3x-x-14+9=0

= x2-5x+x-5 = 0

=x(x-5)+1(x-5)=0

=(x-5) (x+l)-0

Either x-5-0 or x+1=0

Hence, x=5 or x=-1

Therefore, the roots of the given quadratic equation are 5 and -1.

 

Question 19: Find the roots of the equation \(\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}\).

Solution:

We have given with the equation :\(\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}\)

= \(\frac{2x(x-3)+(2x-5)(x-4)}{(x-4)(x-3)}=\frac{25}{3}\)

= \(\frac{2x^{2}-6x+2x^{2}-5x-8x+20}{x^{2}-4x-3x+12}=\frac{25}{3}\)

=  \(\frac{4x^{2}-19x+20}{x^{2}-7x+12}=\frac{25}{3}\)

= 3(4x2-19x+20) = 25(x2-7x+12)

= 12x2-57x+60 = 25x2 – 175x+300

= 13x2-78x-40x+240=0

= 13x2-118x+240=0

= 13x2-78x-40x+240=0

= 13x(x-6)-40(x-6) =0

= (x-6)(13x-40) =0

Either x-6 = 0; x= 6

Or , 13x-40 = 0;  x = 40/13

Therefore, the roots of the given quadratic equation are x=6 and 40/13, respectively.

 

Question 20: Find the roots of the quadratic equation \(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\).

Solution:

We have given with the equation :\(\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}\)

= \(\frac{x(x+3)-(x-2)(1-x)}{x(x-2)}=\frac{17}{4}\)

= \(\frac{x^{2}+3x-x+x^{2}+2-2x}{x^{2}-2x}=\frac{17}{4}\)

= \(\frac{2x^{2}+2}{x^{2}-2x}=\frac{17}{4}\)

= 4(2x2+2) = 17(x2-2x)

= 8x2+8 = 17x2-34x

= 9x2-34x-8 = 0

= 9x2-36x+2x-8 = 0

= 9x(x-4)+2(x-4) =0

= (9x+2)(x-4) =0

Either 9x+2 = 0; x=-2/9

Or, x-4 = 0; x= 4

Therefore, the roots of the given quadratic equation are x=-2/9 and 4 respectively.

 

Question 21: Find the roots of the quadratic equation \(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}\).

Solution:

We have given with the equation :\(\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}\)

= \(\frac{(x-1)+2(x-2)}{(x-2)(x-1)}=\frac{6}{x}\)

= \(\frac{(x-1)+2x-4}{(x^{2}-2x-x+2)}=\frac{6}{x}\)

= \(\frac{3x-5}{(x^{2}-3x+2)}=\frac{6}{x}\)

= x(3x-5) = 6(x2-3x+2)

= 3x2-5x= 6x2-18x+12

= 3x2-13x+12 =0

= 3x2-9x-4x+12 =0

= 3x(x-3)-4(x-3) =0

= (x-3)(3x-4) = 0

Either x-3 = 0; x= 3

Or, 3x-4 = 0; x = 4/3

Therefore, the roots of the given quadratic equation are 3 and \(\frac{4}{3}\) respectively.

 

Question 22: Find the roots of the quadratic equation \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}\).

Solution:

The given equation is: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}\)

= \(\frac{(x+1)^{2}-(x-1)^{2}}{x^{2}-1}=\frac{5}{6}\)

= \(\frac{4x}{x^{2}-1}=\frac{5}{6}\)

= 6(4x) = 5(x2-1)

= 24x= 5x2-5

= 5x2-24x-5 =0

= 5x2-25x+x-5 =0

= 5x(x-5)+1(x-5) =0

= (5x+1)(x-5) =0

Either x-5 = 0 Or 5x+1 = 0

Hence, x= 5 or x=-1/5

Therefore, the roots of the given quadratic equation are\(x=\frac{-1}{5}\) and 5 respectively.

 

Question 23: Find the roots of the quadratic equation \(\frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2}\).

Solution:

The given equation is: \(\frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2}\)

=  \(\frac{(x-1)^{2}+(2x+1)^{2}}{2x^{2}-2x+x-1}=\frac{5}{2} \)

= \(\frac{x^{2}-2x+1+4x^{2}+4x+1}{2x^{2}-x-1}=\frac{5}{2} \)

= \(\frac{5x^{2}+2x+2}{2x^{2}-x-1}=\frac{5}{2} \)

= 2(5x2+2x+2) = 5(2x2-x-1)

= 10x2+4x+4 = 10x2-5x-5

Cancel the equal terms on both sides of the equation.

= 4x+5x+4+5=0

= 9x+9=0

= 9x = -9

x = -1

Hence, x = -1 is the only root we can get, of the given equation.

 

Question 24: Find the roots of the quadratic equation \(\frac{m}{n}x^{2}+\frac{n}{m}=1-2x\)

Solution:

We have given with the equation :\(\frac{m}{n}x^{2}+\frac{n}{m}=1-2x\)

= \(\frac{m}{n}x^{2}+\frac{n}{m}=1-2x\)

= \(\frac{m^{2}x^{2}+n^{2}}{mn}=1-2x\)

= m2x2+2mnx+(n2-mn) = 0

Now we have to solve the above equation using factorization method, therefore;

= \((m^{2}x^{2}+mnx+m\sqrt{mnx})+(mnx-m\sqrt{mnx}(n+\sqrt{mn})(n-\sqrt{mn}))=0\)

= \((m^{2}x^{2}+mnx+m\sqrt{mnx})+(mx(n-\sqrt{mn})+(n+\sqrt{mn})(n-\sqrt{mn}))=0\)

= \(mx(mx+n+\sqrt{mn})+(n-\sqrt{mn})(mx+n+\sqrt{mn})=0\)

= \((mx+n+\sqrt{mn})(mx+n-\sqrt{mn})=0\)

Now, to find the value of x, we will equate both the products to zero. Therefore,

\((mx+n+\sqrt{mn})=0\)

\(mx=-n-\sqrt{mn}\)

\(x=\frac{-n-\sqrt{mn}}{m}\)

Or

\((mx+n-\sqrt{mn})=0\)

\(x=\frac{-n+\sqrt{mn}}{m}\)

\(x=\frac{-n+\sqrt{mn}}{m}\)

Therefore, the roots of the given quadratic equation are\(x=\frac{-n+\sqrt{mn}}{m}\) and \(x=\frac{-n-\sqrt{mn}}{m}\) respectively.

 

Question 25: Find the roots of the quadratic equation \(\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}\)

Solution:

We have given with the equation :\(\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}\)

= \(\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}\)

= \(\frac{(x-a)^{2}+(x-b)^{2}}{(x-a)(x-b)}=\frac{a}{b}+\frac{b}{a}\)

= \(\frac{x^{2}-2ax+a^{2}+x^{2}-2bx+b^{2}}{x^{2}+ab-bx-ax}=\frac{a^{2}+b^{2}}{ab}\)

= (2x2-2x(a+b)+a2+b2)ab = (a2+b2)(x2-(a+b)x+ab)

= (2abx2-2abx(a+b)+ab(a2+b2)) = (a2+b2)(x2-(a2+b2)(a+b)x+(a2+b2)(ab)

= (a2+b2-2ab)x-(a+b)(a2+b2-2ab)x=0

= (a-b)2x2-(a+b)(a+b)2x2=0

= x(a-b)2(x-(a+b))=0

= x(x-(a+b))=0

Either x = 0

Or, (x-(a+b))=0; x= a+b

Therefore, the roots of the given quadratic equation are 0 and a+b respectively.

 

Question 26: Find the roots of the quadratic equation \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\).

Solution:

We have given with the equation :\(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)(x-4)+(x-1)(x-4)+(x-1)(x-2)}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)(x-4)+(x-1)[(x-4)+(x-2)]}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)(x-4)+(x-1)(2x-6)}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)(x-4)+(x-1)2(x-3)}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)[(x-4)+(2x-2)]}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{(x-3)(3x-6)}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

= \(\frac{3(x-3)(x-2)}{(x-1)(x-2)(x-3)(x-3)(x-4)}=\frac{1}{6}\)

Cancel the like terms of numerator and denominator, on both the sides . We get;

= \(\frac{3}{(x-1)(x-2)(x-4)}=\frac{1}{6}\)

= (x-1)(x-4) = 18

= x2-4x-x+4 =18

= x2-5x-14=0

= x2-7x+2x-14=0

= x(x-7)+2(x-7)=0

= (x-7)(x+2)=0

Either x-7 = 0 Or x+2 =0

Therefore, x=7 or x= -2

Therefore, the roots of the given quadratic equation are 7 and -2 respectively.

 

Question 27: Find the roots of the quadratic equation \(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}\).

Solution:

We have given with the equation :\(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}\)

= \(\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}\)

= \(\frac{a(x-b)+b(x-a)}{(x-b)(x-a)}=\frac{2c}{x-c}\)

= \(\frac{ax-ab+bx-ab}{(x^{2}-bx-ax+ab)}=\frac{2c}{x-c}\)

= (x-c)(ax-2ab+bx) = 2c(x2-bx-ax+ab)

= (a+b)x2-2abx-(a+b)cx+2abc= 2cx2-2c(a+b)x+2abc

 

Question 28: Find the roots of the Question x2+2ab=(2a+b)x.

Solution:

We have given with the equation : x2+2ab=(2a+b)x

=> x2+2ab = (2a+b)x

=> x2-(2a+b)x+2ab = 0

=> x2-2ax-bx+2ab = 0

=> x(x-2a)-b(x-2a) =0

=> (x-2a)(x-b) =0

Either x-2a = 0 Or x-b =0

Hence, x= 2a and x= b

Therefore, the roots of the given quadratic equation are 2a and b respectively.

 

Question 29: Find the roots of the quadratic equation (a+b)2x2-4abx-(a-b)2 =0.

Solution:

We have given with the equation : (a+b)2x2-4abx-(a-b)2 =0

=> (a+b)2x2-4abx-(a-b)2 =0

=> (a+b)2x2-((a+b)2 –(a-b)2 )x-(a-b)2 =0

=> (a+b)2x2-(a+b)2 x +(a-b)2 x-(a-b)2 =0

=> (a+b)2x(x-1) (a+b)2 (x-1)=0

=> (x-1) (a+b)2x+(a+b)2) =0

Either x-1 =0 Or (a+b)2x+(a+b)2) =0

Hence, x= 1 and x = \(-(\frac{a-b}{a+b})^{2}\)

Therefore, the roots of the given quadratic equation are\(-(\frac{a-b}{a+b})^{2}\) and 1 respectively .

 

Question 30: Find the roots of the quadratic equation a(x2+1)-x(a2+1)= 0.

Solution:

We have given with the equation : a(x2+1)-x(a2+1)= 0

= a(x2+1)-x(a2+1)= 0

= ax2+a-a2x-x= 0

= ax(x-a)-1(x-a) =0

= (x-a)(ax-1) =0

Either x-a =0 Or ax-1 =0

Hence, x= a or  x=1/a

Therefore, the roots of the given quadratic equation are ( a) and  \(x=\frac{1}{a}\) respectively.

 

Question 31: Find the roots of the quadratic equation \(x^{2}+(a+\frac{1}{a})x+1=0\).

Solution:

We have given with the equation :\(x^{2}+(a+\frac{1}{a})x+1=0\)

= \(x^{2}+(a+\frac{1}{a})x+1=0\)

= \(x^{2}+ax+\frac{x}{a}+(a\times \frac{1}{a})=0\)

= \(x(x+a)+\frac{1}{a}(x+a)=0\)

= \((x+a)(x+\frac{1}{a})=0\)

Either x+a = 0 Or , \((x+\frac{1}{a})=0\)

Hence, x= -a or x = 1/a

Therefore, the roots of the given quadratic equation are\(x=\frac{1}{a}\) and –a respectively.

 

Question 32: Find the roots of the quadratic equation abx2+(b2-ac)x-bc =0

Solution:

We have given with the equation : abx2+(b2-ac)x-bc =0

= abx2+(b2-ac)x-bc =0

= abx2+b2x-acx-bc =0

= bx (ax+b)-c (ax+b)=0

= (ax+b)(bx-c) = 0

Either, ax+b = 0 Or bx-c = 0

Hence, x=-b/a or x=c/b

Therefore, the roots of the given quadratic equation are \(x=\frac{c}{b}\) and \(x=\frac{-b}{a}\) respectively.

 

Question 33: Find the roots of the quadratic equation a2b2x2+b2x-a2x-1=0

Solution:

We have given with the equation : a2b2x2+b2x-a2x-1=0

= a2b2x2+b2x-a2x-1=0

= b2x(a2x+1)-1(a2x+1)

= (a2x+1)( b2x-1) =0

Either (a2x+1) =0 Or ( b2x-1) =0

Hence, x=-1/aor x = 1/b2

Therefore, the roots of the given quadratic equation are \(x=\frac{1}{b^{2}}\) and \(x=\frac{-1}{a^{2}}\) respectively.

 

 

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