Question

Solve the following quadratic equation by factorization :
$(a+b{)}^2{x}^2-4abx-(a-b{)}^2=0$

The roots are 1, $-{\left(\frac{a-b}{a+b}\right)}^2$

• A. True
• B. False

Answer 1

The correct option is A True

$(a+b{)}^2{x}^2-4abx-(a-b{)}^2=0$

As, $-(a+b{)}^2+(a-b{)}^2=-a^2-b^2-2ab+a^2+b^2-2ab=-4ab$

$\Rightarrow$ $(a+b{)}^2{x}^2-(a+b{)}^{2}x+(a-b{)}^{2}x-(a-b{)}^2=0$

$\Rightarrow$ $(a+b{)}^{2}x[x-1]+(a-b{)}^2[x-1]=0$

$\Rightarrow$ $(x-1)\left[\right(a+b{)}^{2}x+(a-b{)}^2]=0$

$\Rightarrow$ $x-1=0$ and $(a+b{)}^{2}x+(a-b{)}^2=0$

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

$\therefore$ $x=1$ and $x=-{\left(\frac{a-b}{a+b}\right)}^2$

$\therefore$ We can see, given roots in question are correct.