How do you use the graph to solve 0=x^2+3x+2?

Jan 22, 2017

The solution is $x = - 1$ and $x = - 2$.

Explanation:

For this, we have to draw the graph first. As

${x}^{2} + 3 x + 2 = {x}^{2} + 2 \times \frac{3}{2} \times x + {\left(\frac{3}{2}\right)}^{2} + 2 - \frac{9}{4} = {\left(x + \frac{3}{2}\right)}^{2} - \frac{1}{4}$

As ${\left(x + \frac{3}{2}\right)}^{2}$ is positive and its minimum value is zero, minimum value of ${x}^{2} + 3 x + 2$ is $- \frac{1}{4}$, at $x = - \frac{3}{2}$.

Let $y = {x}^{2} + 3 x + 2$ and we have a minimum at $\left(- \frac{3}{2} , - \frac{1}{4}\right)$, some other points, we have by putting various values of $x$ (around $x = - \frac{3}{2}$), are

$\left(- 5 , 12\right) , \left(- 4 , 6\right) , \left(- 3 , 2\right) , \left(- 2 , 0\right) , \left(- 1 , 0\right) , \left(0 , 2\right) , \left(1 , 6\right) , \left(2 , 12\right)$ and joining all these eight points and the minimum at $\left(- \frac{3}{2} , - \frac{1}{4}\right)$, we get following graph.
graph{x^2+3x+2 [-4.333, 0.667, -0.43, 2.07]}

As the value of $y = 0$ at $x = - 1$ and $x = - 2$,

the solution is $x = - 1$ and $x = - 2$.