Solve the following quadratic equation by factorization:

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

Given:

Given quadratic equation is $(a+b)^2x^2-4abx-(a-b)^2=0$.

To do:

We have to solve the given quadratic equation.

Solution:

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

To factorise $(a+b)^2x^2-4abx-(a-b)^2=0$, we have to find two numbers $m$ and $n$ such that $m+n=-4ab$ and $mn=(a+b)^2\times(-(a-b)^2)=-((a+b)(a-b))^2=-(a^2-b^2)^2$.

If $m=(a-b)^2$ and $n=-(a+b)^2$, $m+n=a^2+b^2-2ab-(a^2+b^2+2ab)=-4ab$ and $mn=(a-b)^2\times(-(a+b)^2)=-((a+b)(a-b))^2=-(a^2-b^2)^2$.

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

$(a+b)^2x(x-1)+(a-b)^2(x-1)=0$

$((a+b)^2x+(a-b)^2)(x-1)=0$

$(a+b)^2x+(a-b)^2=0$ or $x-1=0$

$(a+b)^2x=-(a-b)^2$ or $x=1$

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

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

The values of $x$ are $-(\frac{a-b}{a+b})^2$ and $1$. 

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Updated on: 10-Oct-2022

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