Question

Solve for x in terms of a and b:

x^2 - 3ax - 4a^2 = 0


x = 4a, x = -a. <<<this is the answer but what are the steps to get it?

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=(-(-3)+/-√(25))/(2)=(3 +/- 5)/(2) \\\\ x_1=(3+5)/(2)=(8)/(2)\to\boxed{4} \\\\ x_2=(3-5)/(2)=(-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}`