How do you solve $x^2-6x-5=0$ by completing the square?

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Answer 1 · 2018-05-31T18:46:58

$x=3+sqrt(14)$ or $x=3-sqrt(14)$

Writing
$x^2-2*3x+9-5-9$
This is
$(x-3)^2=14$
$|x-3|=sqrt(14)$
so we get
$x=3+sqrt(14)$
in the other case

$-x+3=sqrt(14)$
$x=3-sqrt(14)$

Answer 2 · 2018-05-31T18:59:31

$x = +-14+3$

$x = +sqrt14 +3 =6.74 or x = -0.74$

$x^2 -6x-5=0$
$x^2 -6x" "=5" "larr$ move the constant to the right side.

Complete the square by adding $color(red)((-6/2)^2)$ to both sides

$x^2 -6xcolor(red)(+9) =5 color(red)(+9)$

$" "(x-3)^2 = 14" "larr$ write as the square of a binomial

$" "x-3 = +-sqrt14" "larr$ square root both sides.

$x = +-14+3" "larr$ isolate $x$

This gives two solutions for $x:$

$x = +sqrt14 +3 =6.74 or x = -0.74$

How do you solve x^2-6x-5=0x^2-6x-5=0 by completing the square?