Question

Solve each equation. Check your solutions.

$x^2-5x=14$

Short answer

The solution of$x^2-5x=14$ is role="math" localid="1647953707701" $\mathit{x}\mathbf{=}\mathbf{7}\mathbf{,}\mathbf{-}\mathbf{2}$.

Step-by-step solution

Step 1. Subtract.

Subtract from both sides of the given equation.

$$\begin{gathered} x^2-5x-14=14-14 \\ =0 \end{gathered}$$

Step 2. Factor by using GCF.

Factor by using GCF to get the form $xy=0$.

$\left(x-7\right)\left(x+2\right)=0$

Step 3. Write zero product property.

For any real numbers $x$ and $y$, if $xy=0$ then $x=0,y=0$ or both $x$ and $y$ equal to zero.

Step 4. Apply zero product property.

Apply zero product property to the given equation, $\left(x-7\right)\left(x+2\right)=0$.

localid="1647954254435" $$\begin{gathered} x-7=0\text{or}x+2=0 \\ x=7x=-2 \end{gathered}$$

Therefore, the roots are 7 and $-2$.

Step 5. Check the obtained solution.

To check whether the obtained solution is correct or not, substitute 7 for$x$and$-2$for$x$in the given equation and simplify.

role="math" localid="1647954142827" $$\begin{gathered} {\left(7\right)}^2-5\left(7\right)\stackrel{?}{=}14 \\ 49-35\stackrel{?}{=}14 \\ 14=14 \\ LHS=RHS \end{gathered}$$

$$\begin{gathered} {\left(-2\right)}^2-5\left(-2\right)\stackrel{?}{=}14 \\ 4+10\stackrel{?}{=}14 \\ 14=14 \\ LHS=RHS \end{gathered}$$

Therefore, the obtained solution is correct.