Question

Find the zeros of y = x + 6x- 4 by completing the square.

Answer 1

Answer:

$\large\boxed{x=-3\pm\sqrt{13}}$

Step-by-step explanation:

$(a+b)^2=a^2+2ab+b^2\qquad(*)\\\\\\x^2+6x-4=0\qquad\text{add 4 to both sides}\\\\x^2+2(x)(3)=4\qquad\text{add}\ 3^2=9\ \text{to both sides}\\\\\underbrace{x^2+2(x)(3)+3^2}_{(*)}=4+9\\\\(x+3)^2=13\Rightarrow x+3=\pm\sqrt{13}\qquad\text{subtract 3 from both sides}\\\\x=-3\pm\sqrt{13}$