Question
Solve the following equation: $( a + b )^2 x ^2-4 abx -( a - b )^2=0$
✓
Answer
$(a+b)^2 x^2-4 a b x-(a-b)^2=0 $
$ \text { As, }-(a+b)^2+(a-b)^2=-a^2-b^2-2 a b+a^2+b^2-2 a b=-4 a b$
$ (a+b)^2 x^2-\left[(a+b)^2-(a-b)^2\right] x-(a-b)^2=0$
$ (a+b)^2 x^2-(a+b)^2 x+(a-b)^2 x-(a-b)^2=0 $
$ \left\{(a+b)^2 x\right\}(x-1)+\left\{(a-b)^2\right\}(x-1)=0 $
$(x-1)\left[(a+b)^2 x+(a-b)^2\right]=0 $
$ x-1=0 \text { and }(a+b)^2 x+(a-b)^2=0$
$ x=1 \text { and } x=-\frac{(a-b)^2}{(a+b)^2}$
$ x=1 \text { and } x=-\left(\frac{a-b}{a+b}\right)^2$
$ \text { As, }-(a+b)^2+(a-b)^2=-a^2-b^2-2 a b+a^2+b^2-2 a b=-4 a b$
$ (a+b)^2 x^2-\left[(a+b)^2-(a-b)^2\right] x-(a-b)^2=0$
$ (a+b)^2 x^2-(a+b)^2 x+(a-b)^2 x-(a-b)^2=0 $
$ \left\{(a+b)^2 x\right\}(x-1)+\left\{(a-b)^2\right\}(x-1)=0 $
$(x-1)\left[(a+b)^2 x+(a-b)^2\right]=0 $
$ x-1=0 \text { and }(a+b)^2 x+(a-b)^2=0$
$ x=1 \text { and } x=-\frac{(a-b)^2}{(a+b)^2}$
$ x=1 \text { and } x=-\left(\frac{a-b}{a+b}\right)^2$
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