Question

Solve for x in terms of a and b:

Answer 1

$x^2 - 3ax - 4a^2 = 0\\ x^2+ax-4ax-4a^2=0\\ x(x+a)-4a(x+a)=0\\ (x-4a)(x+a)=0\\ x=4a \vee x=-a$

Answer 2

$x^2-3ax-4a^2=0 \\\\ a=1 \\ b=-3 \\ c=-4 \\\\ \Delta=(-3)^2-1*(-4)*(-4)=9+16 \to \boxed{25} \\\\ x_1;x_2=\frac{-(-3)+/-\sqrt{25}}{2}=\frac{3 +/- 5}{2} \\\\ x_1=\frac{3+5}{2}=\frac{8}{2}\to\boxed{4} \\\\ x_2=\frac{3-5}{2}=\frac{-2}{2}\to\boxed{-1} \\\\ x^2-3ax-4a^2=0 \\\\ (x-4a)(x-(-1)*a)=0 \\\\ 1)x-4a=0 \ => \boxed{x=4a} \\\\ 2)x+a=0 \ => \boxed{x=a}$