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

Answer:

The solutions to the quadratic equations are:

[tex]x=\sqrt{13}+3,\:x=-\sqrt{13}+3[/tex]

Step-by-step explanation:

Given the function

[tex]y\:=\:x^2\:-\:6x\:-\:4[/tex]

substitute y = 0 in the equation to determine the zeros

[tex]0\:=\:x^2\:-\:6x\:-\:4[/tex]

Switch sides

[tex]x^2-6x-4=0[/tex]

Add 4 to both sides

[tex]x^2-6x-4+4=0+4[/tex]

Simplify

[tex]x^2-6x=4[/tex]

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

[tex]x^2-6x+\left(-3\right)^2=4+\left(-3\right)^2[/tex]

[tex]x^2-6x+\left(-3\right)^2=13[/tex]

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

[tex]\left(x-3\right)^2=13[/tex]

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

solve

[tex]x-3=\sqrt{13}[/tex]

Add 3 to both sides

[tex]x-3+3=\sqrt{13}+3[/tex]

Simplify

[tex]x=\sqrt{13}+3[/tex]

now solving

[tex]x-3=-\sqrt{13}[/tex]

Add 3 to both sides

[tex]x-3+3=-\sqrt{13}+3[/tex]

Simplify

[tex]x=-\sqrt{13}+3[/tex]

Thus, the solutions to the quadratic equations are:

[tex]x=\sqrt{13}+3,\:x=-\sqrt{13}+3[/tex]

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