Question

Solve the given quadratic equation by completing the square, $t^2+3t=7$

Answer 1

The given quadratic equation is $t^2+3t=7$ which can be rewritten as $t^2+3t-7=0$.

Add and subtract $\frac{9}{4}$ in the equation $t^2+3t-7=0$ to make a perfect square as shown below:

$t^2-3t-7=0\Rightarrow t^2-3t+\frac{9}{4}-\frac{9}{4}-7=0\Rightarrow (t{)}^2+(2\times t\times \frac{3}{2})+{\left(\frac{3}{2}\right)}^2+(-\frac{9}{4}-7)=0\Rightarrow {(t+\frac{3}{2})}^2-\frac{37}{4}=0(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow {(t+\frac{3}{2})}^2=\frac{37}{4}\Rightarrow t+\frac{3}{2}=\pm \sqrt{\frac{37}{4}}\Rightarrow t+\frac{3}{2}=\pm \frac{\sqrt{37}}{2}\Rightarrow t+\frac{3}{2}=-\frac{\sqrt{37}}{2},t+\frac{3}{2}=\frac{\sqrt{37}}{2}\Rightarrow t=-\frac{\sqrt{37}}{2}-\frac{3}{2},t=\frac{\sqrt{37}}{2}-\frac{3}{2}\Rightarrow t=-\left(\frac{\sqrt{37}+3}{2}\right),t=\frac{\sqrt{37}-3}{2}$

Hence, $t=-\left(\frac{\sqrt{37}+3}{2}\right)$ or $t=\frac{\sqrt{37}-3}{2}$.