Question

In the following exercises, solve by completing the square.

$2p^2+7p=14$

Short answer

The solution is$p=-\frac{\sqrt{161}}{4}-\frac{7}{4},p=\frac{\sqrt{161}}{4}-\frac{7}{4}$

Step-by-step solution

Step 1. Given information 

Equation:$2p^2+7p=14$

Step 2. Calculation    

We have,

$2p^2+7p=14$

$\frac{2p^2+7p}{2}=\frac{14}{2}$

$p^2+\frac{7p}{2}+{\left(\frac{1}{2}·\left(\frac{7}{2}\right)\right)}^2=7+{\left(\frac{1}{2}·\left(\frac{7}{2}\right)\right)}^2$

$p^2+\frac{7p}{2}+{\left(\frac{7}{4}\right)}^2=7+{\left(\frac{7}{4}\right)}^2$

$p^2+\frac{7p}{2}+\frac{49}{16}=7+\frac{49}{16}$

${\left(p+\frac{7}{4}\right)}^2=\frac{161}{16}$

$\sqrt{{\left(p+\frac{7}{4}\right)}^2}=\sqrt{\frac{161}{16}}$

$p+\frac{7}{4}=\pm \frac{\sqrt{161}}{4}$

$p+\frac{7}{4}-\frac{7}{4}=-\frac{\sqrt{161}}{4}-\frac{7}{4},p+\frac{7}{4}-\frac{7}{4}=\frac{\sqrt{161}}{4}-\frac{7}{4}$

localid="1645180152491" $p=-\frac{\sqrt{161}}{4}-\frac{7}{4},p=\frac{\sqrt{161}}{4}-\frac{7}{4}$