Question:
Short-Answer Questions
Solve: $x^{2}-4 a x+4 a^{2}-b^{2}=0$
Solution:
$x^{2}-4 a x+4 a^{2}-b^{2}=0$
$\Rightarrow x^{2}-4 a x+(2 a+b)(2 a-b)=0$
$\Rightarrow x^{2}-[(2 a+b)+(2 a-b)] x+(2 a+b)(2 a-b)=0$
$\Rightarrow x^{2}-(2 a+b) x-(2 a-b) x+(2 a+b)(2 a-b)=0$
$\Rightarrow x[x-(2 a+b)]-(2 a-b)[x-(2 a+b)]=0$
$\Rightarrow[x-(2 a+b)][x-(2 a-b)]=0$
$\Rightarrow x-(2 a+b)=0$ or $x-(2 a-b)=0$
$\Rightarrow x=(2 a+b)$ or $x=(2 a-b)$
Hence, (2a + b) and (2a − b) are the roots of the given equation.
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