Question

Solve: $x^2-4ax+4a^2-b^2=0.$

Answer 1

$$\begin{gathered} x^2-4ax+4a^2-b^2=0 \\ \text{Here,} \\ a=1,b=-4a\mathrm{and}c=4a^2-b^2 \\ \therefore D=\sqrt{b^2-4ac} \\ =\sqrt{16a^2-4(4a^2-b^2)} \\ \text{=}\sqrt{16a^2-16a^2+4b^2} \\ \text{= 2}b \\ \therefore \text{The roots of the equation are:} \\ \frac{-b+D}{2a},\frac{-b-D}{2a} \\ \mathrm{or},\frac{4a+2b}{2},\frac{4a-2b}{2} \\ \mathrm{or},\frac{\sout{2}(2a+b)}{\sout{2}},\frac{\sout{2}(2a-b)}{\sout{2}} \\ \mathrm{or},\text{(}2a+b),(2a-b) \end{gathered}$$