Question
Solve the following quadratic equations by factorization:
$x^2+2 a b=(2 a+b) x$
$x^2+2 a b=(2 a+b) x$
✓
Answer
We have
$x^2+2 a b=(2 a+b) x$
$\Rightarrow x^2-(2 a+b) x+2 a b=0$
${[\because 2 a b=-8 a x-b \Rightarrow-(8 a+b)=-8 a-b}$
$\Rightarrow x^2-2 a x-b x+2 a b=0$
$\Rightarrow x-(x-8 a)-b(x-2 a)=0$
$\Rightarrow(x-8 a)(x-b)=0$
$\Rightarrow x-8 a=0 \text { or } x-b=0$
$\Rightarrow x=8 a=0 \text { or } x=b$
$\therefore x = 8a$ and $x = b$ are the two roots of the given equation.
$x^2+2 a b=(2 a+b) x$
$\Rightarrow x^2-(2 a+b) x+2 a b=0$
${[\because 2 a b=-8 a x-b \Rightarrow-(8 a+b)=-8 a-b}$
$\Rightarrow x^2-2 a x-b x+2 a b=0$
$\Rightarrow x-(x-8 a)-b(x-2 a)=0$
$\Rightarrow(x-8 a)(x-b)=0$
$\Rightarrow x-8 a=0 \text { or } x-b=0$
$\Rightarrow x=8 a=0 \text { or } x=b$
$\therefore x = 8a$ and $x = b$ are the two roots of the given equation.
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