# How do you solve and write the following in interval notation: 4x^2 + x ≤ 3?

Jun 10, 2016

Solution is $- 1 \le x \le \frac{3}{4}$.

#### Explanation:

$4 {x}^{2} + x \le 3$ can be written as $4 {x}^{2} + x - 3 \le 0$

Now factorizing LHS, we get

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

or $4 x \left(x + 1\right) - 3 \left(x + 1\right) \le 0$

or $\left(4 x - 3\right) \left(x + 1\right) \le 0$

As equality sign is fulfilled by $x = - \frac{3}{4}$ and $x = - 1$ and these are part of solution as we already have equality sign.

These two points divides number line in three regions

A- First region is $x < - 1$ - In this region both $\left(4 x - 3\right)$ and $\left(x + 1\right)$ are negative and as such $4 {x}^{2} + x - 3 > 0$, hence this is not a solution.

B- Second region is $- 1 < x < \frac{3}{4}$ - In this region while $\left(x + 1\right)$ is positive, $\left(4 x - 3\right)$ is negative and as such $4 {x}^{2} + x - 3 < 0$, hence this is a solution.

C- Third region is $\frac{3}{4} < x$ - In this region while $\left(x + 1\right)$ and $\left(4 x - 3\right)$ both are positive and as such $4 {x}^{2} + x - 3 > 0$, hence this is a not a solution.

Hence solution is $- 1 \le x \le \frac{3}{4}$.