Solve sqrt{x+10}-sqrt{15-x}=1

$$\sqrt { x + 10 } - \sqrt { 15 - x } = 1$$

Solve for x

$$x=6$$

Solution Steps

Subtract -\sqrt{15-x} from both sides of the equation.

$$\sqrt{x+10}=1+\sqrt{15-x}$$

Square both sides of the equation.

$$\left(\sqrt{x+10}\right)^{2}=\left(1+\sqrt{15-x}\right)^{2}$$

Calculate \sqrt{x+10} to the power of 2 and get x+10.

$$x+10=\left(1+\sqrt{15-x}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{15-x}\right)^{2}.

$$x+10=1+2\sqrt{15-x}+\left(\sqrt{15-x}\right)^{2}$$

Calculate \sqrt{15-x} to the power of 2 and get 15-x.

$$x+10=1+2\sqrt{15-x}+15-x$$

Add 1 and 15 to get 16.

$$x+10=16+2\sqrt{15-x}-x$$

Subtract 16-x from both sides of the equation.

$$x+10-\left(16-x\right)=2\sqrt{15-x}$$

To find the opposite of 16-x, find the opposite of each term.

$$x+10-16+x=2\sqrt{15-x}$$

Subtract 16 from 10 to get -6.

$$x-6+x=2\sqrt{15-x}$$

Combine x and x to get 2x.

$$2x-6=2\sqrt{15-x}$$

Square both sides of the equation.

$$\left(2x-6\right)^{2}=\left(2\sqrt{15-x}\right)^{2}$$

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

$$4x^{2}-24x+36=\left(2\sqrt{15-x}\right)^{2}$$

Expand \left(2\sqrt{15-x}\right)^{2}.

$$4x^{2}-24x+36=2^{2}\left(\sqrt{15-x}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-24x+36=4\left(\sqrt{15-x}\right)^{2}$$

Calculate \sqrt{15-x} to the power of 2 and get 15-x.

$$4x^{2}-24x+36=4\left(15-x\right)$$

Use the distributive property to multiply 4 by 15-x.

$$4x^{2}-24x+36=60-4x$$

Subtract 60 from both sides.

$$4x^{2}-24x+36-60=-4x$$

Subtract 60 from 36 to get -24.

$$4x^{2}-24x-24=-4x$$

Add 4x to both sides.

$$4x^{2}-24x-24+4x=0$$

Combine -24x and 4x to get -20x.

$$4x^{2}-20x-24=0$$

Divide both sides by 4.

$$x^{2}-5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=1\left(-6\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=-6 b=1$$

Rewrite x^{2}-5x-6 as \left(x^{2}-6x\right)+\left(x-6\right).

$$\left(x^{2}-6x\right)+\left(x-6\right)$$

Factor out x in x^{2}-6x.

$$x\left(x-6\right)+x-6$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x+1\right)$$

To find equation solutions, solve x-6=0 and x+1=0.

$$x=6 x=-1$$

Substitute 6 for x in the equation \sqrt{x+10}-\sqrt{15-x}=1.

$$\sqrt{6+10}-\sqrt{15-6}=1$$

Simplify. The value x=6 satisfies the equation.

$$1=1$$

Substitute -1 for x in the equation \sqrt{x+10}-\sqrt{15-x}=1.

$$\sqrt{-1+10}-\sqrt{15-\left(-1\right)}=1$$

Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute 6 for x in the equation \sqrt{x+10}-\sqrt{15-x}=1.

$$\sqrt{6+10}-\sqrt{15-6}=1$$

Simplify. The value x=6 satisfies the equation.

$$1=1$$

Equation \sqrt{x+10}=\sqrt{15-x}+1 has a unique solution.

$$x=6$$