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Solve the following quadratic equation by factorization:
$4x^2\ +\ 4bx\ –\ (a^2\ –\ b^2)\ =\ 0$
Given:
Given quadratic equation is $4x^2\ +\ 4bx\ –\ (a^2\ –\ b^2)\ =\ 0$.
To do:
We have to solve the given quadratic equation by factorization.
Solution:
To factorise $4x^2\ +\ 4bx\ –\ (a^2\ –\ b^2)\ =\ 0$, we have to find two numbers $m$ and $n$ such that $m+n=4b$ and $mn=4\times(-(a^2-b^2))=-4(a+b)(a-b)$ ($a^2-b^2=(a+b)(a-b)$)
If $m=-2(a-b)$ and $n=2(a+b)$,
$m+n=-2(a-b)+2(a+b)=-2a+2b+2a+2b=4b$
$mn=-2(a-b)\times2(a+b)=-4(a^2-b^2)$
Therefore,
$4x^2+4bx-(a^2-b^2)=0$
$4x^2-2(a-b)x+2(a+b)x-(a-b)(a+b)=0$
$2x(2x-(a-b))+(a+b)(2x-(a-b))=0$
$(2x-(a-b))(2x+(a+b))=0$
$2x-(a-b)=0$ or $2x+(a+b)=0$
$2x=a-b$ or $2x=-(a+b)$
$x=\frac{a-b}{2}$ or $x=\frac{-(a+b)}{2}$
The roots of the given quadratic equation are $\frac{a-b}{2}$ and $\frac{-(a+b)}{2}$.