completing the square to solve equation

Question

Is it possible to use the method of completing the square to solve the equation $2x^2+18x+21=0$ ?

I have problem with how to remove the negative sign on the right side.

Answer 1

Hint: $2x^2 + 18x + 21 = 2(x^2 + 9x) + 21 = 2\left(x+\dfrac{9}{2}\right)^2 - \dfrac{81}{2} + 21$.

Answer 2

$$ 2\left[\left(x+\frac92\right)^2+\frac{21}{2}-\frac{81}{4}\right]= 2\left[\left(x+\frac92\right)^2-\frac{39}{4}\right] $$ and so on.

Answer 3

First make the leading coefficient $1$:

$x^2 + 9x + \frac{21}{2} = 0$.

Then add and subtract $(\frac{9}{2})^2$:

$x^2 + 9x + \frac{21}{2} + (\frac{9}{2})^2 - (\frac{9}{2})^2 = 0$

Rearrange,

$x^2 + 9x + (\frac{9}{2})^2 = (\frac{9}{2})^2 - \frac{21}{2}$

Observe the LHS is a square:

$(x + \frac 92)^2 = \frac{81}{4} - \frac{42}{4} = \frac{39}{4}$

Take square root of both sides,

$x + \frac 92 = \pm \frac 12 \sqrt {39}$

Final solution:

$x = - \frac 92 \pm \frac 12 \sqrt {39}$

Answer 4

Make the leading coefficient a square avoiding fractions by multiplying $2x^2+18x=-21$ by $2$, you'll get $4x^2+36x=-42$. Now add $81$ to complete the square: $$4x^2+36x+81=39\iff(2x+9)^2=39\iff2x+9=\pm\sqrt{39}\iff x=\frac{-9\pm\sqrt39}{2}.$$

Answer 5

Here is a graphical solution that starts with completing the square and finishes off with a difference of squares. I hope it downloads okay for everyone.