Question

In the following exercises, solve by completing the square.

$3q^2-5q=9$

Short answer

The solution is$q=-\frac{\sqrt{133}}{6}+\frac{5}{6},q=\frac{\sqrt{133}}{6}+\frac{5}{6}$

Step-by-step solution

Step 1. Given information 

Equation:$3q^2-5q=9$

Step 2. Calculation    

We have,

$3q^2-5q=9$

$\frac{3q^2-5q}{3}=\frac{9}{3}$

$q^2-\frac{5q}{3}+{\left(\frac{1}{2}·\left(-\frac{5}{3}\right)\right)}^2=3+{\left(\frac{1}{2}·\left(-\frac{5}{3}\right)\right)}^2$

role="math" localid="1645180355962" $q^2-\frac{5q}{3}+\frac{25}{36}=3+\frac{25}{36}$

role="math" localid="1645180450868" ${\left(q-\frac{5}{6}\right)}^2=\frac{133}{36}$

$\sqrt{{\left(q-\frac{5}{6}\right)}^2}=\sqrt{\frac{133}{36}}$

$q-\frac{5}{6}=\pm \frac{\sqrt{133}}{6}$

$q-\frac{5}{6}+\frac{5}{6}=-\frac{\sqrt{133}}{6}+\frac{5}{6},q-\frac{5}{6}+\frac{5}{6}=\frac{\sqrt{133}}{6}+\frac{5}{6}$

$q=-\frac{\sqrt{133}}{6}+\frac{5}{6},q=\frac{\sqrt{133}}{6}+\frac{5}{6}$