Solve sqrt{x+6}=5-sqrt{x+1}

$$\sqrt { x + 6 } = 5 - \sqrt { x + 1 }$$

Solve for x

$$x=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+6}\right)^{2}=\left(5-\sqrt{x+1}\right)^{2}$$

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

$$x+6=\left(5-\sqrt{x+1}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x+1}\right)^{2}.

$$x+6=25-10\sqrt{x+1}+\left(\sqrt{x+1}\right)^{2}$$

Calculate \sqrt{x+1} to the power of 2 and get x+1.

$$x+6=25-10\sqrt{x+1}+x+1$$

Add 25 and 1 to get 26.

$$x+6=26-10\sqrt{x+1}+x$$

Add 10\sqrt{x+1} to both sides.

$$x+6+10\sqrt{x+1}=26+x$$

Subtract x from both sides.

$$x+6+10\sqrt{x+1}-x=26$$

Combine x and -x to get 0.

$$6+10\sqrt{x+1}=26$$

Subtract 6 from both sides.

$$10\sqrt{x+1}=26-6$$

Subtract 6 from 26 to get 20.

$$10\sqrt{x+1}=20$$

Divide both sides by 10.

$$\sqrt{x+1}=\frac{20}{10}$$

Divide 20 by 10 to get 2.

$$\sqrt{x+1}=2$$

Square both sides of the equation.

$$x+1=4$$

Subtract 1 from both sides of the equation.

$$x+1-1=4-1$$

Subtracting 1 from itself leaves 0.

$$x=4-1$$

Subtract 1 from 4.

$$x=3$$

Substitute 3 for x in the equation \sqrt{x+6}=5-\sqrt{x+1}.

$$\sqrt{3+6}=5-\sqrt{3+1}$$

Simplify. The value x=3 satisfies the equation.

$$3=3$$

Equation \sqrt{x+6}=-\sqrt{x+1}+5 has a unique solution.

$$x=3$$