Q6. - Algebra 1

Question

Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest tenth.

${\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathit{x}\mathbf{+}\mathbf{10}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The solutions of the equation are $\mathbf{-}\mathbf{5}$ and $\mathbf{-}\mathbf{2}$.

Step 1. Graph the equation.

Graph the equation $y=x^2+7x+10$ to find the roots.

Step 2. State the interpretation of the graph.

In the graph of the equation $y=x^2+7x+10$ it can be observed that the x-intercepts appear to be at $-5$ and $-2$, therefore, the solutions are  $-5$ and $-2$ .

Step 3. Check the solutions.

 To check the solutions first substitute $-5$  for $x$ into the equation $y=x^2+7x+10$ and simplify.

$\begin{array}{d}{x}^2+7x+10\stackrel{?}{=}0\\ {(-5)}^2+7(-5)+10\stackrel{?}{=}0\\ 25-35+10\stackrel{?}{=}0\\ 10-10\stackrel{?}{=}0\\ 0\stackrel{?}{=}0\text{ True}\end{array}$

Now substitute $-2$ for $x$ into the equation $y=x^2+7x+10$ and simplify.

$\begin{array}{l}{x}^2+7x+10\stackrel{?}{=}0\\ {\left(-2\right)}^2+7(-2)+10\stackrel{?}{=}0\\ 4-14+10\stackrel{?}{=}0\\ 14-14\stackrel{?}{=}0\\ 0\stackrel{?}{=}0 \text{ True}\end{array}$

Therefore, the solutions of the equation are $-5$ and $-2$.