Solve sqrt{x^2+5^2}=x+1

$$\sqrt { x ^ { 2 } + 5 ^ { 2 } } = x + 1$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+5^{2}}\right)^{2}=\left(x+1\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$\left(\sqrt{x^{2}+25}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{x^{2}+25} to the power of 2 and get x^{2}+25.

$$x^{2}+25=\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}+25=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$x^{2}+25-x^{2}=2x+1$$

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

$$25=2x+1$$

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

$$2x+1=25$$

Subtract 1 from both sides.

$$2x=25-1$$

Subtract 1 from 25 to get 24.

$$2x=24$$

Divide both sides by 2.

$$x=\frac{24}{2}$$

Divide 24 by 2 to get 12.

$$x=12$$

Substitute 12 for x in the equation \sqrt{x^{2}+5^{2}}=x+1.

$$\sqrt{12^{2}+5^{2}}=12+1$$

Simplify. The value x=12 satisfies the equation.

$$13=13$$

Equation \sqrt{x^{2}+25}=x+1 has a unique solution.

$$x=12$$