Question

Solve the equation both algebraically and graphically. (Lesson 1.11)

$x^3+16=0$

Short answer

The solution by algebraically is $x\approx 2.52$and graphically is given below:

Step-by-step solution

Step 1. Definition of the intersection point.

An intersection is a single point where two lines meet or cross each other.

Step 2. Solve equation algebraically.

We have$x^3+16=0$

Subtract 16 both sides:

$\begin{array}{l}{x}^3+16-16=0-16\\ x^3=-16\end{array}$

Take cubic root to both sides:

$\begin{array}{l}x=\sqrt[3]{-16}\\ x=\sqrt[3]{-8\cdot 2}\\ x=-2\sqrt[3]{2}\\ x\approx -2.52\end{array}$

Step 3. Solve the equation graphically.

$x^3+16=0$

Now, equate the left side of the equation to$y_1$and the right side to $y_2$.

$\begin{array}{l}{y}_1=x^3+16\\ y_2=0\end{array}$

Draw the graph:

As we can see that both lines are intersecting on $\left(-2.52,0\right)$.

So, the solution is $x=-2.52$.