# How do you solve the quadratic equation by completing the square: 4x^2 + 9 = 12x?

Jul 15, 2015

$x = \frac{3}{2}$ is a "double root" (a root of multiplicity 2 ).

#### Explanation:

First, rearrange so the ${x}^{2}$ and $x$ terms are on the left and the constant term is on the right: $4 {x}^{2} - 12 x = - 9$.

Next, factor out the coefficient of ${x}^{2}$ on the left: $4 \left({x}^{2} - 3 x\right) = - 9$.

Next, you can either divide both sides by 4 before completing the square or complete the square first. Let's complete the square on ${x}^{2} - 3 x$ first. Take the coefficient of $x$, which is $- 3$, divide it by 2 to get $- \frac{3}{2}$, then square this number to get $\frac{9}{4}$. Add that number inside the parentheses and balance on the right side of the equation to get:

$4 \left({x}^{2} - 3 x + \frac{9}{4}\right) = - 9 + 4 \cdot \frac{9}{4} = 0$ (it doesn't usually come out to be zero on the right here, that's just a coincidence).

Once you've done this procedure, the expression inside the parentheses will be a perfect square. In this case, we can write:

$4 {\left(x - \frac{3}{2}\right)}^{2} = 0$

In general, we can now divide both sides by the coefficient in front, which is 4 in this case. Since $\frac{0}{4} = 0$, we get ${\left(x - \frac{3}{2}\right)}^{2} = 0$.

Now take the $\setminus \pm$ square root of both sides. Since $\setminus \pm \sqrt{0} = 0$, there is only one equation and one root of the original equation: $x - \frac{3}{2} = 0$ so $x = \frac{3}{2}$ is a root of multiplicity 2.