4.6 Completing the square

$x^2=\text{18}$

$x^2=0$

$x^2=-\text{60}$

$x^2+\mathrm{8x}+\text{12}=0$

• A

  Solve by factoring.

• B

  Now, we’re going to solve it a different way. Start by adding four to both sides.

• C

  Now, the left side can be written as $(x+\text{something}{)}^2$ . Rewrite it that way, and then solve from there.

• D

  Did you get the same answers this way that you got by factoring?

$(x-3{)}^2=x^2-\mathrm{6x}+\text{\_\_\_}$

$(x-\frac{3}{2}{)}^2=x^2-\text{\_\_\_}x+\text{\_\_\_}$

$(x+\text{\_\_\_}{)}^2=x^2+\text{10}x+\text{\_\_\_}$

$(x-\text{\_\_\_}{)}^2=x^2-\text{18}x+\text{\_\_\_}$

$x^2-\text{20}x+\text{90}=\text{26}$

${\mathrm{3x}}^2+\mathrm{2x}-4=0$