How do you solve the quadratic equation by completing the square: $x^2-6x-8=0$ ?

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Answer 1 · 2015-07-14T23:00:50

$x=3+sqrt(17), 3-sqrt(17)$

$x^2-6x-8=0$

In order to solve a quadratic equation by completing the square, we must force a perfect square trinomial on the left side. $a^2+2ab+b^2=(a+b)^2$

Add $8$ to both sides of the equation.

$x^2-6x=8$

Divide the coefficient of the $x$ term by $2$ and square the result. Add it to both sides of the equation.

$((-6)/(2))^2=(-3)^2=9$

$x^2-6x+9=8+9$ =

$x^2-6x+9=17$

We now have a perfect square trinomial on the left side, in which
$a=x,$ and $b=-3$ .

Now we can solve for $x$ .

$(x-3)^2=17$

Take the square root of both sides.

$(x-3)=+-sqrt(17)$

$x=3+-sqrt(17)$

$x=3+sqrt(17)$

$x=3-sqrt(17)$

How do you solve the quadratic equation by completing the square: x^2-6x-8=0x^2-6x-8=0?