How do you solve the quadratic equation by completing the square: $x^2+4x=5$ ?

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Answer 1 · 2015-07-19T12:04:09

Complete the square by adding $4$ to both sides of the equation to find:

$(x+2)^2 = x^2+4x+4 = 5+4 = 9 = 3^2$

Hence $x = 1$ or $x=-5$

Given $ax^2+bx$ , note that

$a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)$

In our case $a = 1$ and $b = 4$ , so $b/(2a) = 4/2 = 2$ and

$(x+2)^2 = x^2+4x+4$

So add $4$ to both sides of our original equation to get:

$x^2+4x+4 = 9$

That is:

$(x+2)^2 = 3^2$

So

$x+2 = +-sqrt(3^2) = +-3$

Subtract $2$ from both sides to get:

$x = -2 +-3$

How do you solve the quadratic equation by completing the square: x^2+4x=5x^2+4x=5?