Solve sqrt{left(-4-0right)^2+left(1-0right)^2+left(7-zright)^2}=sqrt{left(3-0right)^2+{left(5-0right)}^2+{left(-2-zright)}^2}

$$\sqrt{ { \left(-4-0 \right) }^{ 2 } + { \left(1-0 \right) }^{ 2 } + { \left(7-z \right) }^{ 2 } } = \sqrt{ { \left(3-0 \right) }^{ 2 } + { \left(5-0 \right) }^{ 2 } + { \left(-2-z \right) }^{ 2 } }$$

Solve for z

$$z = \frac{14}{9} = 1\frac{5}{9} \approx 1.555555556$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(-4-0\right)^{2}+\left(1-0\right)^{2}+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Subtract 0 from -4 to get -4.

$$\left(\sqrt{\left(-4\right)^{2}+\left(1-0\right)^{2}+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$\left(\sqrt{16+\left(1-0\right)^{2}+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Subtract 0 from 1 to get 1.

$$\left(\sqrt{16+1^{2}+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Calculate 1 to the power of 2 and get 1.

$$\left(\sqrt{16+1+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Add 16 and 1 to get 17.

$$\left(\sqrt{17+\left(7-z\right)^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7-z\right)^{2}.

$$\left(\sqrt{17+49-14z+z^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Add 17 and 49 to get 66.

$$\left(\sqrt{66-14z+z^{2}}\right)^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Calculate \sqrt{66-14z+z^{2}} to the power of 2 and get 66-14z+z^{2}.

$$66-14z+z^{2}=\left(\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Subtract 0 from 3 to get 3.

$$66-14z+z^{2}=\left(\sqrt{3^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$66-14z+z^{2}=\left(\sqrt{9+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Subtract 0 from 5 to get 5.

$$66-14z+z^{2}=\left(\sqrt{9+5^{2}+\left(-2-z\right)^{2}}\right)^{2}$$

Calculate 5 to the power of 2 and get 25.

$$66-14z+z^{2}=\left(\sqrt{9+25+\left(-2-z\right)^{2}}\right)^{2}$$

Add 9 and 25 to get 34.

$$66-14z+z^{2}=\left(\sqrt{34+\left(-2-z\right)^{2}}\right)^{2}$$

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

$$66-14z+z^{2}=\left(\sqrt{34+4+4z+z^{2}}\right)^{2}$$

Add 34 and 4 to get 38.

$$66-14z+z^{2}=\left(\sqrt{38+4z+z^{2}}\right)^{2}$$

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

$$66-14z+z^{2}=38+4z+z^{2}$$

Subtract 4z from both sides.

$$66-14z+z^{2}-4z=38+z^{2}$$

Combine -14z and -4z to get -18z.

$$66-18z+z^{2}=38+z^{2}$$

Subtract z^{2} from both sides.

$$66-18z+z^{2}-z^{2}=38$$

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

$$66-18z=38$$

Subtract 66 from both sides.

$$-18z=38-66$$

Subtract 66 from 38 to get -28.

$$-18z=-28$$

Divide both sides by -18.

$$z=\frac{-28}{-18}$$

Reduce the fraction \frac{-28}{-18} to lowest terms by extracting and canceling out -2.

$$z=\frac{14}{9}$$

Substitute \frac{14}{9} for z in the equation \sqrt{\left(-4-0\right)^{2}+\left(1-0\right)^{2}+\left(7-z\right)^{2}}=\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-z\right)^{2}}.

$$\sqrt{\left(-4-0\right)^{2}+\left(1-0\right)^{2}+\left(7-\frac{14}{9}\right)^{2}}=\sqrt{\left(3-0\right)^{2}+\left(5-0\right)^{2}+\left(-2-\frac{14}{9}\right)^{2}}$$

Simplify. The value z=\frac{14}{9} satisfies the equation.

$$\frac{1}{9}\times 3778^{\frac{1}{2}}=\frac{1}{9}\times 3778^{\frac{1}{2}}$$

Equation \sqrt{\left(7-z\right)^{2}+17}=\sqrt{\left(-z-2\right)^{2}+34} has a unique solution.

$$z=\frac{14}{9}$$