Solve sqrt{17+(1-x)^2}=sqrt{34+(-2-x)^2}

$$\sqrt { 17 + ( 1 - x ) ^ { 2 } } = \sqrt { 34 + ( - 2 - x ) ^ { 2 } }$$

Solve for x

$$x = -\frac{10}{3} = -3\frac{1}{3} \approx -3.333333333$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{17+\left(1-x\right)^{2}}\right)^{2}=\left(\sqrt{34+\left(-2-x\right)^{2}}\right)^{2}$$

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

$$\left(\sqrt{17+1-2x+x^{2}}\right)^{2}=\left(\sqrt{34+\left(-2-x\right)^{2}}\right)^{2}$$

Add 17 and 1 to get 18.

$$\left(\sqrt{18-2x+x^{2}}\right)^{2}=\left(\sqrt{34+\left(-2-x\right)^{2}}\right)^{2}$$

Calculate \sqrt{18-2x+x^{2}} to the power of 2 and get 18-2x+x^{2}.

$$18-2x+x^{2}=\left(\sqrt{34+\left(-2-x\right)^{2}}\right)^{2}$$

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

$$18-2x+x^{2}=\left(\sqrt{34+4+4x+x^{2}}\right)^{2}$$

Add 34 and 4 to get 38.

$$18-2x+x^{2}=\left(\sqrt{38+4x+x^{2}}\right)^{2}$$

Calculate \sqrt{38+4x+x^{2}} to the power of 2 and get 38+4x+x^{2}.

$$18-2x+x^{2}=38+4x+x^{2}$$

Subtract 4x from both sides.

$$18-2x+x^{2}-4x=38+x^{2}$$

Combine -2x and -4x to get -6x.

$$18-6x+x^{2}=38+x^{2}$$

Subtract x^{2} from both sides.

$$18-6x+x^{2}-x^{2}=38$$

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

$$18-6x=38$$

Subtract 18 from both sides.

$$-6x=38-18$$

Subtract 18 from 38 to get 20.

$$-6x=20$$

Divide both sides by -6.

$$x=\frac{20}{-6}$$

Reduce the fraction \frac{20}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{10}{3}$$

Substitute -\frac{10}{3} for x in the equation \sqrt{17+\left(1-x\right)^{2}}=\sqrt{34+\left(-2-x\right)^{2}}.

$$\sqrt{17+\left(1-\left(-\frac{10}{3}\right)\right)^{2}}=\sqrt{34+\left(-2-\left(-\frac{10}{3}\right)\right)^{2}}$$

Simplify. The value x=-\frac{10}{3} satisfies the equation.

$$\frac{1}{3}\times 322^{\frac{1}{2}}=\frac{1}{3}\times 322^{\frac{1}{2}}$$

Equation \sqrt{\left(1-x\right)^{2}+17}=\sqrt{\left(-x-2\right)^{2}+34} has a unique solution.

$$x=-\frac{10}{3}$$