Solve sqrt{x-11}=5-sqrt{x+4}

$$\sqrt { x - 11 } = 5 - \sqrt { x + 4 }$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x-11}\right)^{2}=\left(5-\sqrt{x+4}\right)^{2}$$

Calculate \sqrt{x-11} to the power of 2 and get x-11.

$$x-11=\left(5-\sqrt{x+4}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x+4}\right)^{2}.

$$x-11=25-10\sqrt{x+4}+\left(\sqrt{x+4}\right)^{2}$$

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

$$x-11=25-10\sqrt{x+4}+x+4$$

Add 25 and 4 to get 29.

$$x-11=29-10\sqrt{x+4}+x$$

Add 10\sqrt{x+4} to both sides.

$$x-11+10\sqrt{x+4}=29+x$$

Subtract x from both sides.

$$x-11+10\sqrt{x+4}-x=29$$

Combine x and -x to get 0.

$$-11+10\sqrt{x+4}=29$$

Add 11 to both sides.

$$10\sqrt{x+4}=29+11$$

Add 29 and 11 to get 40.

$$10\sqrt{x+4}=40$$

Divide both sides by 10.

$$\sqrt{x+4}=\frac{40}{10}$$

Divide 40 by 10 to get 4.

$$\sqrt{x+4}=4$$

Square both sides of the equation.

$$x+4=16$$

Subtract 4 from both sides of the equation.

$$x+4-4=16-4$$

Subtracting 4 from itself leaves 0.

$$x=16-4$$

Subtract 4 from 16.

$$x=12$$

Substitute 12 for x in the equation \sqrt{x-11}=5-\sqrt{x+4}.

$$\sqrt{12-11}=5-\sqrt{12+4}$$

Simplify. The value x=12 satisfies the equation.

$$1=1$$

Equation \sqrt{x-11}=-\sqrt{x+4}+5 has a unique solution.

$$x=12$$