(a) Solve the quadratic equation for x: x^2-2 a x-(4 b^2-a^2)=0

\[x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0 \] $\displaystyle x^{2}-2 a x-\left(4 b^{2}-a^{2}\right)=0$ gives \[\begin{array}{l} x^{2}-2 a x+a^{2}-4 b^{2}=0 \\ \Rightarrow(x-a)^{2}-(2 b)^{2}=0 \\ \Rightarrow(x-a+2 b)(x-a-2 b)=0 \\ \therefore x=a-2 b \text { and } a+2 b \end{array} \]Or (b) ``` If the quadratic equation ($\displaystyle 1$ + a $\displaystyle 2$ ) x2 + 2abx + (b $\displaystyle 2$ - c2) = $\displaystyle 0$ has equal and real roots, them prove that: b2 = c2 ($\displaystyle 1$ + a $\displaystyle 2$) ```The equation has equal roots therefore $\displaystyle 4 a^{2} b^{2}-4\left(1+a^{2}\right)\left(b^{2}-c^{2}\right)=0$ \[\Rightarrow b^{2}=c^{2}\left(1+a^{2}\right) \]