How do you solve $x^{2} + 2 x = 9$ $x^2+2x=9$ by completing the square?

Question

4987 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-12T07:40:11

The solutions are $S = {- 4.162, 2.162}$ $S={-4.162, 2.162}$

Let's complete the square

$x^{2} + 2 x = 9$ $x^2+2x=9$

We add ${(\frac{2}{2})}^{2}$ $(2/2)^2$ to both sides

$x^{2} + 2 x + 1 = 9 + 1$ $x^2+2x+1=9+1$

The $L H S$ $LHS$ is

$(a^{2} + 2 a b + b^{2}) = {(a + b)}^{2}$ $(a^2+2ab+b^2)=(a+b)^2$

${(x + 1)}^{2} = 10$ $(x+1)^2=10$

Taking square roots on both sides

$x + 1 = \pm \sqrt{10}$ $x+1=+-sqrt10$

$x = - 1 \pm \sqrt{10}$ $x=-1+-sqrt10$

$x_{1} = - 1 - \sqrt{10} = - 4.162$ $x_1=-1-sqrt10=-4.162$

$x_{2} = - 1 + \sqrt{10} = 2.162$ $x_2=-1+sqrt10=2.162$

How do you solve x^2+2x=9x2+2x=9x^2+2x=9 by completing the square?