# How do I use a sign chart to solve 2x^2-7x<=-3?

Jun 30, 2015

2x^2 – 7x ≤ -3 when 0.5 ≤ x ≤3

#### Explanation:

Let's modify the inequality by adding $3$ to each side. Then

2x^2 – 7x + 3 ≤ 0

We start by finding the critical numbers.

Set f(x) = 2x^2 – 7x + 3 = 0 and solve for $x$.

$\left(2 x - 1\right) \left(x - 3\right) = 0$

$2 x - 1 = 0$ or $x - 3 = 0$

$2 x = 1$

x = ½ = 0.5 or $x = 3$

The critical numbers are $0.5$ and $3$.

Now we check for positive and negative regions.

We have three regions to consider: (a) x ≤ 0.5; (b) -3 ≤ x ≤ 0.5; and (c) x ≥ 0.5.

Case (a): Let $x = 0$.

Then f(0) = 2×0^2 - 7×0 + 3 =0 - 0 + 3 = 3

f(x) ≥ 0 when x ≤ 0.5.

Case (b): Let $x = 1$.

Then f(1) = 2×1^2 - 7×1 + 3 = 2 - 7 + 3 = -2

f(x) ≤ 0 when 0.5 ≤ x ≤ 3

Case (c): Let $x = 4$.

Then f(1) = 2×4^2 - 7×4 + 3 = 32 - 28 + 3 = 7

f(x) ≥ 0 when x ≥ 3.

If 2x^2 – 7x + 3 ≤ 0 when 0.5 ≤ x ≤ 3, then

2x^2 – 7x ≤ -3 when 0.5 ≤ x ≤ 3