Solve sqrt{a^2+2^2}=sqrt{(a+3)^2+1}

$$\sqrt { a ^ { 2 } + 2 ^ { 2 } } = \sqrt { ( a + 3 ) ^ { 2 } + 1 }$$

Solve for a

$$a=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{a^{2}+2^{2}}\right)^{2}=\left(\sqrt{\left(a+3\right)^{2}+1}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$\left(\sqrt{a^{2}+4}\right)^{2}=\left(\sqrt{\left(a+3\right)^{2}+1}\right)^{2}$$

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

$$a^{2}+4=\left(\sqrt{\left(a+3\right)^{2}+1}\right)^{2}$$

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

$$a^{2}+4=\left(\sqrt{a^{2}+6a+9+1}\right)^{2}$$

Add 9 and 1 to get 10.

$$a^{2}+4=\left(\sqrt{a^{2}+6a+10}\right)^{2}$$

Calculate \sqrt{a^{2}+6a+10} to the power of 2 and get a^{2}+6a+10.

$$a^{2}+4=a^{2}+6a+10$$

Subtract a^{2} from both sides.

$$a^{2}+4-a^{2}=6a+10$$

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

$$4=6a+10$$

Swap sides so that all variable terms are on the left hand side.

$$6a+10=4$$

Subtract 10 from both sides.

$$6a=4-10$$

Subtract 10 from 4 to get -6.

$$6a=-6$$

Divide both sides by 6.

$$a=\frac{-6}{6}$$

Divide -6 by 6 to get -1.

$$a=-1$$

Substitute -1 for a in the equation \sqrt{a^{2}+2^{2}}=\sqrt{\left(a+3\right)^{2}+1}.

$$\sqrt{\left(-1\right)^{2}+2^{2}}=\sqrt{\left(-1+3\right)^{2}+1}$$

Simplify. The value a=-1 satisfies the equation.

$$5^{\frac{1}{2}}=5^{\frac{1}{2}}$$

Equation \sqrt{a^{2}+4}=\sqrt{\left(a+3\right)^{2}+1} has a unique solution.

$$a=-1$$