Solve sqrt{9-a}=3+a

$$\sqrt { 9 - a } = 3 + a$$

Solve for a

$$a=0$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{9-a}\right)^{2}=\left(3+a\right)^{2}$$

Calculate \sqrt{9-a} to the power of 2 and get 9-a.

$$9-a=\left(3+a\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+a\right)^{2}.

$$9-a=9+6a+a^{2}$$

Subtract 9 from both sides.

$$9-a-9=6a+a^{2}$$

Subtract 9 from 9 to get 0.

$$-a=6a+a^{2}$$

Subtract 6a from both sides.

$$-a-6a=a^{2}$$

Combine -a and -6a to get -7a.

$$-7a=a^{2}$$

Subtract a^{2} from both sides.

$$-7a-a^{2}=0$$

Factor out a.

$$a\left(-7-a\right)=0$$

To find equation solutions, solve a=0 and -7-a=0.

$$a=0 a=-7$$

Substitute 0 for a in the equation \sqrt{9-a}=3+a.

$$\sqrt{9-0}=3+0$$

Simplify. The value a=0 satisfies the equation.

$$3=3$$

Substitute -7 for a in the equation \sqrt{9-a}=3+a.

$$\sqrt{9-\left(-7\right)}=3-7$$

Simplify. The value a=-7 does not satisfy the equation because the left and the right hand side have opposite signs.

$$4=-4$$

Equation \sqrt{9-a}=a+3 has a unique solution.

$$a=0$$