Solve sqrt{t^2+7t+4}=t+9

$$\sqrt { t ^ { 2 } + 7 t + 4 } = t + 9$$

Solve for t

$$t=-7$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{t^{2}+7t+4}\right)^{2}=\left(t+9\right)^{2}$$

Calculate \sqrt{t^{2}+7t+4} to the power of 2 and get t^{2}+7t+4.

$$t^{2}+7t+4=\left(t+9\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(t+9\right)^{2}.

$$t^{2}+7t+4=t^{2}+18t+81$$

Subtract t^{2} from both sides.

$$t^{2}+7t+4-t^{2}=18t+81$$

Combine t^{2} and -t^{2} to get 0.

$$7t+4=18t+81$$

Subtract 18t from both sides.

$$7t+4-18t=81$$

Combine 7t and -18t to get -11t.

$$-11t+4=81$$

Subtract 4 from both sides.

$$-11t=81-4$$

Subtract 4 from 81 to get 77.

$$-11t=77$$

Divide both sides by -11.

$$t=\frac{77}{-11}$$

Divide 77 by -11 to get -7.

$$t=-7$$

Substitute -7 for t in the equation \sqrt{t^{2}+7t+4}=t+9.

$$\sqrt{\left(-7\right)^{2}+7\left(-7\right)+4}=-7+9$$

Simplify. The value t=-7 satisfies the equation.

$$2=2$$

Equation \sqrt{t^{2}+7t+4}=t+9 has a unique solution.

$$t=-7$$