# How do you solve 2x^2-7x+3>=0 by graphing?

Feb 22, 2017

$x \le \frac{1}{2} \text{ or } x \ge 3$

#### Explanation:

Factorise the quadratic and find the roots.

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

$\Rightarrow \text{roots are "x=1/2" and } x = 3$

$x = 0 \to y = 3 \Rightarrow \left(0 , 3\right)$

$\text{coefficient of " x^2>0" thus} \cup$

With this information we can sketch the function.
graph{2x^2-7x+3 [-10, 10, -5, 5]}

To solve $2 {x}^{2} + 7 x - 3 \ge 0$ Consider the region above and including the x-axis.

Above x-axis to the left of $\frac{1}{2}$ and right of 3

$\Rightarrow x \le \frac{1}{2} \text{ or } x \ge 3$