# How do you solve and write the following in interval notation: 4x²+12x+9≤0?

Jul 21, 2016

$x = - \frac{3}{2}$

#### Explanation:

Note that as $4 {x}^{2} + 12 x + 9 = {\left(2 x\right)}^{2} + 2 \times 2 \times x 3 + {3}^{2}$, its factors using identity ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$ are ${\left(2 x + 3\right)}^{2}$.

As ${\left(2 x + 3\right)}^{2}$ is a perfect square, it cannot be negative and the only solution is

when $4 {x}^{2} + 12 x + 9 = 0$ or ${\left(2 x + 3\right)}^{2} = 0$ i.e.

$2 x + 3 = 0$ or $x = - \frac{3}{2}$

graph{4x^2+12x+9 [-2.402, 0.098, -0.28, 0.97]}