Solve sqrt{x^2+4x+13}+sqrt{x^2-2x-2}=5

$$\sqrt { x ^ { 2 } + 4 x + 13 } + \sqrt { x ^ { 2 } - 2 x - 2 } = 5$$

Solve for x

$$x=\frac{5-5\sqrt{13}}{8}\approx -1.628469547$$

Solution Steps

Subtract \sqrt{x^{2}-2x-2} from both sides of the equation.

$$\sqrt{x^{2}+4x+13}=5-\sqrt{x^{2}-2x-2}$$

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+4x+13} to the power of 2 and get x^{2}+4x+13.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x^{2}-2x-2}\right)^{2}.

$$x^{2}+4x+13=25-10\sqrt{x^{2}-2x-2}+\left(\sqrt{x^{2}-2x-2}\right)^{2}$$

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

$$x^{2}+4x+13=25-10\sqrt{x^{2}-2x-2}+x^{2}-2x-2$$

Subtract 2 from 25 to get 23.

$$x^{2}+4x+13=23-10\sqrt{x^{2}-2x-2}+x^{2}-2x$$

Subtract 23+x^{2}-2x from both sides of the equation.

$$x^{2}+4x+13-\left(23+x^{2}-2x\right)=-10\sqrt{x^{2}-2x-2}$$

To find the opposite of 23+x^{2}-2x, find the opposite of each term.

$$x^{2}+4x+13-23-x^{2}+2x=-10\sqrt{x^{2}-2x-2}$$

Subtract 23 from 13 to get -10.

$$x^{2}+4x-10-x^{2}+2x=-10\sqrt{x^{2}-2x-2}$$

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

$$4x-10+2x=-10\sqrt{x^{2}-2x-2}$$

Combine 4x and 2x to get 6x.

$$6x-10=-10\sqrt{x^{2}-2x-2}$$

Square both sides of the equation.

$$\left(6x-10\right)^{2}=\left(-10\sqrt{x^{2}-2x-2}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-10\right)^{2}.

$$36x^{2}-120x+100=\left(-10\sqrt{x^{2}-2x-2}\right)^{2}$$

Expand \left(-10\sqrt{x^{2}-2x-2}\right)^{2}.

$$36x^{2}-120x+100=\left(-10\right)^{2}\left(\sqrt{x^{2}-2x-2}\right)^{2}$$

Calculate -10 to the power of 2 and get 100.

$$36x^{2}-120x+100=100\left(\sqrt{x^{2}-2x-2}\right)^{2}$$

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

$$36x^{2}-120x+100=100\left(x^{2}-2x-2\right)$$

Use the distributive property to multiply 100 by x^{2}-2x-2.

$$36x^{2}-120x+100=100x^{2}-200x-200$$

Subtract 100x^{2} from both sides.

$$36x^{2}-120x+100-100x^{2}=-200x-200$$

Combine 36x^{2} and -100x^{2} to get -64x^{2}.

$$-64x^{2}-120x+100=-200x-200$$

Add 200x to both sides.

$$-64x^{2}-120x+100+200x=-200$$

Combine -120x and 200x to get 80x.

$$-64x^{2}+80x+100=-200$$

Add 200 to both sides.

$$-64x^{2}+80x+100+200=0$$

Add 100 and 200 to get 300.

$$-64x^{2}+80x+300=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -64 for a, 80 for b, and 300 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-80±\sqrt{80^{2}-4\left(-64\right)\times 300}}{2\left(-64\right)}$$

Square 80.

$$x=\frac{-80±\sqrt{6400-4\left(-64\right)\times 300}}{2\left(-64\right)}$$

Multiply -4 times -64.

$$x=\frac{-80±\sqrt{6400+256\times 300}}{2\left(-64\right)}$$

Multiply 256 times 300.

$$x=\frac{-80±\sqrt{6400+76800}}{2\left(-64\right)}$$

Add 6400 to 76800.

$$x=\frac{-80±\sqrt{83200}}{2\left(-64\right)}$$

Take the square root of 83200.

$$x=\frac{-80±80\sqrt{13}}{2\left(-64\right)}$$

Multiply 2 times -64.

$$x=\frac{-80±80\sqrt{13}}{-128}$$

Now solve the equation x=\frac{-80±80\sqrt{13}}{-128} when ± is plus. Add -80 to 80\sqrt{13}.

$$x=\frac{80\sqrt{13}-80}{-128}$$

Divide -80+80\sqrt{13} by -128.

$$x=\frac{5-5\sqrt{13}}{8}$$

Now solve the equation x=\frac{-80±80\sqrt{13}}{-128} when ± is minus. Subtract 80\sqrt{13} from -80.

$$x=\frac{-80\sqrt{13}-80}{-128}$$

Divide -80-80\sqrt{13} by -128.

$$x=\frac{5\sqrt{13}+5}{8}$$

The equation is now solved.

$$x=\frac{5-5\sqrt{13}}{8} x=\frac{5\sqrt{13}+5}{8}$$

Substitute \frac{5-5\sqrt{13}}{8} for x in the equation \sqrt{x^{2}+4x+13}+\sqrt{x^{2}-2x-2}=5.

$$\sqrt{\left(\frac{5-5\sqrt{13}}{8}\right)^{2}+4\times \frac{5-5\sqrt{13}}{8}+13}+\sqrt{\left(\frac{5-5\sqrt{13}}{8}\right)^{2}-2\times \frac{5-5\sqrt{13}}{8}-2}=5$$

Simplify. The value x=\frac{5-5\sqrt{13}}{8} satisfies the equation.

$$5=5$$

Substitute \frac{5\sqrt{13}+5}{8} for x in the equation \sqrt{x^{2}+4x+13}+\sqrt{x^{2}-2x-2}=5.

$$\sqrt{\left(\frac{5\sqrt{13}+5}{8}\right)^{2}+4\times \frac{5\sqrt{13}+5}{8}+13}+\sqrt{\left(\frac{5\sqrt{13}+5}{8}\right)^{2}-2\times \frac{5\sqrt{13}+5}{8}-2}=5$$

Simplify. The value x=\frac{5\sqrt{13}+5}{8} does not satisfy the equation.

$$\frac{15}{4}+\frac{3}{4}\times 13^{\frac{1}{2}}=5$$

Substitute \frac{5-5\sqrt{13}}{8} for x in the equation \sqrt{x^{2}+4x+13}+\sqrt{x^{2}-2x-2}=5.

$$\sqrt{\left(\frac{5-5\sqrt{13}}{8}\right)^{2}+4\times \frac{5-5\sqrt{13}}{8}+13}+\sqrt{\left(\frac{5-5\sqrt{13}}{8}\right)^{2}-2\times \frac{5-5\sqrt{13}}{8}-2}=5$$

Simplify. The value x=\frac{5-5\sqrt{13}}{8} satisfies the equation.

$$5=5$$

Equation \sqrt{x^{2}+4x+13}=-\sqrt{x^{2}-2x-2}+5 has a unique solution.

$$x=\frac{5-5\sqrt{13}}{8}$$