Solve sqrt{x+2}-sqrt{2x-3}=1

$$\sqrt { x + 2 } - \sqrt { 2 x - 3 } = 1$$

Solve for x

$$x=2$$

Solution Steps

Subtract -\sqrt{2x-3} from both sides of the equation.

$$\sqrt{x+2}=1+\sqrt{2x-3}$$

Square both sides of the equation.

$$\left(\sqrt{x+2}\right)^{2}=\left(1+\sqrt{2x-3}\right)^{2}$$

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

$$x+2=\left(1+\sqrt{2x-3}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+\sqrt{2x-3}\right)^{2}.

$$x+2=1+2\sqrt{2x-3}+\left(\sqrt{2x-3}\right)^{2}$$

Calculate \sqrt{2x-3} to the power of 2 and get 2x-3.

$$x+2=1+2\sqrt{2x-3}+2x-3$$

Subtract 3 from 1 to get -2.

$$x+2=-2+2\sqrt{2x-3}+2x$$

Subtract -2+2x from both sides of the equation.

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

To find the opposite of -2+2x, find the opposite of each term.

$$x+2+2-2x=2\sqrt{2x-3}$$

Add 2 and 2 to get 4.

$$x+4-2x=2\sqrt{2x-3}$$

Combine x and -2x to get -x.

$$-x+4=2\sqrt{2x-3}$$

Square both sides of the equation.

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

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

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

Expand \left(2\sqrt{2x-3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$x^{2}-8x+16=4\left(\sqrt{2x-3}\right)^{2}$$

Calculate \sqrt{2x-3} to the power of 2 and get 2x-3.

$$x^{2}-8x+16=4\left(2x-3\right)$$

Use the distributive property to multiply 4 by 2x-3.

$$x^{2}-8x+16=8x-12$$

Subtract 8x from both sides.

$$x^{2}-8x+16-8x=-12$$

Combine -8x and -8x to get -16x.

$$x^{2}-16x+16=-12$$

Add 12 to both sides.

$$x^{2}-16x+16+12=0$$

Add 16 and 12 to get 28.

$$x^{2}-16x+28=0$$

To solve the equation, factor x^{2}-16x+28 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-16 ab=28$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 28.

$$-1,-28 -2,-14 -4,-7$$

Calculate the sum for each pair.

$$-1-28=-29 -2-14=-16 -4-7=-11$$

The solution is the pair that gives sum -16.

$$a=-14 b=-2$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-14\right)\left(x-2\right)$$

To find equation solutions, solve x-14=0 and x-2=0.

$$x=14 x=2$$

Substitute 14 for x in the equation \sqrt{x+2}-\sqrt{2x-3}=1.

$$\sqrt{14+2}-\sqrt{2\times 14-3}=1$$

Simplify. The value x=14 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-1=1$$

Substitute 2 for x in the equation \sqrt{x+2}-\sqrt{2x-3}=1.

$$\sqrt{2+2}-\sqrt{2\times 2-3}=1$$

Simplify. The value x=2 satisfies the equation.

$$1=1$$

Equation \sqrt{x+2}=\sqrt{2x-3}+1 has a unique solution.

$$x=2$$