Question
Solve the following quadratic equation:
$4x^2- 4ax + (a^2 - b^2) = 0$ where $a , b ∈ R$.
$4x^2- 4ax + (a^2 - b^2) = 0$ where $a , b ∈ R$.
✓
Answer
$4 x^2-4 a x+\left(a^2-b^2\right)=0 \text { where } a, b \in R$
$ \Rightarrow 4 x^2-\{2(a+b) x+2(a-b) x\}+a^2-b^2=0 $
$\Rightarrow\left\{4 x^2-2(a+b) x\right\}-\left\{2(a-b) x-\left(a^2-b^2\right)\right\}=0 $
$\Rightarrow 2 x\{2 x-(a+b)\}-(a-b)\{2 x-(a+b)\}=0$
$ \Rightarrow\{2 x-(a+b)\}\{2 x-(a-b)\}=0 $
$ \Rightarrow 2 x-(a+b)=0$
or
$2 x -( a - b )=0$
$\Rightarrow x =\frac{a+b}{2} \text { or } x =\frac{a-b}{2} .$
$ \Rightarrow 4 x^2-\{2(a+b) x+2(a-b) x\}+a^2-b^2=0 $
$\Rightarrow\left\{4 x^2-2(a+b) x\right\}-\left\{2(a-b) x-\left(a^2-b^2\right)\right\}=0 $
$\Rightarrow 2 x\{2 x-(a+b)\}-(a-b)\{2 x-(a+b)\}=0$
$ \Rightarrow\{2 x-(a+b)\}\{2 x-(a-b)\}=0 $
$ \Rightarrow 2 x-(a+b)=0$
or
$2 x -( a - b )=0$
$\Rightarrow x =\frac{a+b}{2} \text { or } x =\frac{a-b}{2} .$
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