Question

Solve the given equation graphically: $x^2+x-12=0$

Answer 1

The given quadratic equation is $x^2+x-12=0$.

Let $y=x^2+x-12$, when we substitute different values of $x$ in the equation $y=x^2+x-12$ then value of $y$ changes accordingly.

When $x=0$ then

$y=(0{)}^2+0-12=0+0-12=-12$

The point is $(0,-12)$

When $x=3$ then

$y=(3{)}^2+3-12=9+3-12=0$

The point is $(3,0)$

When $x=-4$ then

$y=(-4{)}^2-4-12=16-4-12=0$

The point is $(-4,0)$

Therefore, the coordinates are $(0,-12)$, $(3,0)$ and $(-4,0)$ and the graph of the quadratic equation $y=x^2+x-12$ is as shown above.

Hence, from above graph, we get that $x=-4$ or $x=3$.