Question

Solve by completing the square:

$2p^2+7p=14$.

Short answer

The value of $p$for the equation $2p^2+7p=14$by completing the square are$\frac{\sqrt{161}-7}{4},\frac{-\sqrt{161}-7}{4}$.

Step-by-step solution

Step 1. Given Information.

The equation is,

$2p^2+7p=14$.

Step 2. Completing the square.

Given, $2p^2+7p=14$

Dividing both sides by $2$to eliminate the coefficient of $p^2$,

$$\begin{gathered} \Rightarrow \frac{2p^2+7p}{2}=\frac{14}{2} \\ \Rightarrow p^2+\frac{7}{2}p=7 \end{gathered}$$

Adding the square of half of $\frac{7}{2}$on both the sides,

$$\begin{gathered} \Rightarrow p^2+\frac{7}{2}p+{\left(\frac{1}{2}\left(\frac{7}{2}\right)\right)}^2=7+{\left(\frac{1}{2}\left(\frac{7}{2}\right)\right)}^2 \\ \Rightarrow p^2+\frac{7}{2}p+\frac{49}{16}=7+\frac{49}{16} \\ \Rightarrow p^2+\frac{7}{2}p+\frac{49}{16}=\frac{161}{16} \end{gathered}$$

Writing the Left Hand Side as a binomial square.

role="math" localid="1653672948257" $\Rightarrow {\left(p+\frac{7}{4}\right)}^2=\frac{161}{16}$

Taking the square root on both sides,

$$\begin{gathered} \Rightarrow p+\frac{7}{4}=\sqrt{\frac{161}{16}} \\ \Rightarrow p+\frac{7}{4}=\pm \frac{\sqrt{161}}{4} \end{gathered}$$

Solving for the value of $p$,

$$\begin{gathered} \Rightarrow p=\pm \frac{\sqrt{161}}{4}-\frac{7}{4} \\ \Rightarrow p=\frac{\sqrt{161}-7}{4} \\ or,p=\frac{-\sqrt{161}-7}{4} \\ \Rightarrow p=\frac{\sqrt{161}-7}{4},\frac{-\sqrt{161}-7}{4} \end{gathered}$$