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Solve the following quadratic equation for $x: x^{2}-4ax-b^{2}+4a^{2}=0$.
Given: Equation: $x^{2}-4ax-b^{2}+4a^{2}=0$.
To do: To solve the equation for $x$.
Solution:
Given equation: $x^{2}-4ax-b^{2}+4a^{2}=0$.
$\Rightarrow x^2-4ax+4a^2-b^2=0$
$\Rightarrow x^2-2\times2a\times x+(2a)^2-(b)^2=0$
$\Rightarrow (x-2a)^2-(b)^2=0$
$\Rightarrow (x-2a-b)(x-2a+b)=0$
If $x-2a-b=0$
$\Rightarrow x=2a+b$
If $x-2a+b=0$
$\Rightarrow x=2a-b$
Therefore, $x= 2a+b$ or $2a-b$
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