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hoa [83]
3 years ago
13

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

Mathematics
1 answer:
katrin [286]3 years ago
4 0

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

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