Solve sqrt{x^2+3^2}=sqrt{9+(2-x)^2-2*5}

$$\sqrt { x ^ { 2 } + 3 ^ { 2 } } = \sqrt { 9 + ( 2 - x ) ^ { 2 } - 2 \cdot 5 }$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x^{2}+3^{2}}\right)^{2}=\left(\sqrt{9+\left(2-x\right)^{2}-2\times 5}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\left(\sqrt{x^{2}+9}\right)^{2}=\left(\sqrt{9+\left(2-x\right)^{2}-2\times 5}\right)^{2}$$

Calculate \sqrt{x^{2}+9} to the power of 2 and get x^{2}+9.

$$x^{2}+9=\left(\sqrt{9+\left(2-x\right)^{2}-2\times 5}\right)^{2}$$

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

$$x^{2}+9=\left(\sqrt{9+4-4x+x^{2}-2\times 5}\right)^{2}$$

Add 9 and 4 to get 13.

$$x^{2}+9=\left(\sqrt{13-4x+x^{2}-2\times 5}\right)^{2}$$

Multiply 2 and 5 to get 10.

$$x^{2}+9=\left(\sqrt{13-4x+x^{2}-10}\right)^{2}$$

Subtract 10 from 13 to get 3.

$$x^{2}+9=\left(\sqrt{3-4x+x^{2}}\right)^{2}$$

Calculate \sqrt{3-4x+x^{2}} to the power of 2 and get 3-4x+x^{2}.

$$x^{2}+9=3-4x+x^{2}$$

Add 4x to both sides.

$$x^{2}+9+4x=3+x^{2}$$

Subtract x^{2} from both sides.

$$x^{2}+9+4x-x^{2}=3$$

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

$$9+4x=3$$

Subtract 9 from both sides.

$$4x=3-9$$

Subtract 9 from 3 to get -6.

$$4x=-6$$

Divide both sides by 4.

$$x=\frac{-6}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{3}{2}$$

Substitute -\frac{3}{2} for x in the equation \sqrt{x^{2}+3^{2}}=\sqrt{9+\left(2-x\right)^{2}-2\times 5}.

$$\sqrt{\left(-\frac{3}{2}\right)^{2}+3^{2}}=\sqrt{9+\left(2-\left(-\frac{3}{2}\right)\right)^{2}-2\times 5}$$

Simplify. The value x=-\frac{3}{2} satisfies the equation.

$$\frac{3}{2}\times 5^{\frac{1}{2}}=\frac{3}{2}\times 5^{\frac{1}{2}}$$

Equation \sqrt{x^{2}+9}=\sqrt{\left(2-x\right)^{2}+9-10} has a unique solution.

$$x=-\frac{3}{2}$$