Question

Find the zeros of y = x2 – 6x – 4 by completing the square.

Answer 1

Answer:

The solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$

Step-by-step explanation:

Given the function

$y\:=\:x^2\:-\:6x\:-\:4$

substitute y = 0 in the equation to determine the zeros

$0\:=\:x^2\:-\:6x\:-\:4$

Switch sides

$x^2-6x-4=0$

Add 4 to both sides

$x^2-6x-4+4=0+4$

Simplify

$x^2-6x=4$

Rewrite in the form (x+a)² = b

But, in order to rewrite in the form x²+2ax+a²

Solve for 'a'

2ax = -6x

a = -3

so add a² = (-3)² to both sides

$x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2$

$x^2-6x+\left(-3\right)^2=13$

Apply perfect square formula:  (a-b)² = a²-2ab+b²

$\left(x-3\right)^2=13$

$\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

solve

$x-3=\sqrt{13}$

Add 3 to both sides

$x-3+3=\sqrt{13}+3$

Simplify

$x=\sqrt{13}+3$

now solving

$x-3=-\sqrt{13}$

Add 3 to both sides

$x-3+3=-\sqrt{13}+3$

Simplify

$x=-\sqrt{13}+3$

Thus, the solutions to the quadratic equations are:

$x=\sqrt{13}+3,\:x=-\sqrt{13}+3$