Solve sqrt{(-4-x)^2+289-100}=sqrt{(3-x)^2}

$$\sqrt { ( - 4 - x ) ^ { 2 } + 289 - 100 } = \sqrt { ( 3 - x ) ^ { 2 } }$$

Solve for x

$$x=-14$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{\left(-4-x\right)^{2}+289-100}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4-x\right)^{2}.

$$\left(\sqrt{16+8x+x^{2}+289-100}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}}\right)^{2}$$

Add 16 and 289 to get 305.

$$\left(\sqrt{305+8x+x^{2}-100}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}}\right)^{2}$$

Subtract 100 from 305 to get 205.

$$\left(\sqrt{205+8x+x^{2}}\right)^{2}=\left(\sqrt{\left(3-x\right)^{2}}\right)^{2}$$

Calculate \sqrt{205+8x+x^{2}} to the power of 2 and get 205+8x+x^{2}.

$$205+8x+x^{2}=\left(\sqrt{\left(3-x\right)^{2}}\right)^{2}$$

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

$$205+8x+x^{2}=\left(\sqrt{9-6x+x^{2}}\right)^{2}$$

Calculate \sqrt{9-6x+x^{2}} to the power of 2 and get 9-6x+x^{2}.

$$205+8x+x^{2}=9-6x+x^{2}$$

Add 6x to both sides.

$$205+8x+x^{2}+6x=9+x^{2}$$

Combine 8x and 6x to get 14x.

$$205+14x+x^{2}=9+x^{2}$$

Subtract x^{2} from both sides.

$$205+14x+x^{2}-x^{2}=9$$

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

$$205+14x=9$$

Subtract 205 from both sides.

$$14x=9-205$$

Subtract 205 from 9 to get -196.

$$14x=-196$$

Divide both sides by 14.

$$x=\frac{-196}{14}$$

Divide -196 by 14 to get -14.

$$x=-14$$

Substitute -14 for x in the equation \sqrt{\left(-4-x\right)^{2}+289-100}=\sqrt{\left(3-x\right)^{2}}.

$$\sqrt{\left(-4-\left(-14\right)\right)^{2}+289-100}=\sqrt{\left(3-\left(-14\right)\right)^{2}}$$

Simplify. The value x=-14 satisfies the equation.

$$17=17$$

Equation \sqrt{\left(-x-4\right)^{2}+189}=\sqrt{\left(3-x\right)^{2}} has a unique solution.

$$x=-14$$