Solve this

Question:

$x^{2}-2 a x+\left(a^{2}-b^{2}\right)=0$

 

Solution:

Given:

$x^{2}-2 a x+\left(a^{2}-b^{2}\right)=0$

On comparing it with $\mathrm{A} x^{2}+B x+C=0$, we get:

$A=1, B=-2 a$ and $C=\left(a^{2}-b^{2}\right)$

Discriminant $D$ is given by:

$D=B^{2}-4 A C$

$=(-2 a)^{2}-4 \times 1 \times\left(a^{2}-b^{2}\right)$

$=4 a^{2}-4 a^{2}+4 b^{2}$

$=4 b^{2}>0$

Hence, the roots of the equation are real.

Roots $\alpha$ and $\beta$ are given by:

$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-2 a)+\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a+2 b}{2}=\frac{2(a+b)}{2}=(a+b)$

$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-2 a)-\sqrt{4 b^{2}}}{2 \times 1}=\frac{2 a-2 b}{2}=\frac{2(a-b)}{2}=(a-b)$

Hence, the roots of the equation are $(a+b)$ and $(a-b)$.

 

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