Question

find the roots of the equation x2-2(a2+b2)x+(a2-b2)2 = 0

Answer 1

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Answer 2

Given:

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

To find:

Roots of the equation.

Solution:

We can solve the above mathematical problem with the following approach.

We know that (a²-b²) = (a + b) (a - b).

So the given equation can be written as:

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

Therefore, the roots of the equation $x^{2}-2\left(a^{2}+b^{2}\right) x-\left(a^{2}-b^{2}\right)^{2}=0$ are $x=(a+b)^{2},(a-b)^{2}$.