Solve sqrt{2x+5}-sqrt{x-2}=3

$$\sqrt { 2 x + 5 } - \sqrt { x - 2 } = 3$$

Solve for x

$$x=38$$

$$x=2$$

Solution Steps

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

$$\sqrt{2x+5}=3+\sqrt{x-2}$$

Square both sides of the equation.

$$\left(\sqrt{2x+5}\right)^{2}=\left(3+\sqrt{x-2}\right)^{2}$$

Calculate \sqrt{2x+5} to the power of 2 and get 2x+5.

$$2x+5=\left(3+\sqrt{x-2}\right)^{2}$$

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

$$2x+5=9+6\sqrt{x-2}+\left(\sqrt{x-2}\right)^{2}$$

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

$$2x+5=9+6\sqrt{x-2}+x-2$$

Subtract 2 from 9 to get 7.

$$2x+5=7+6\sqrt{x-2}+x$$

Subtract 7+x from both sides of the equation.

$$2x+5-\left(7+x\right)=6\sqrt{x-2}$$

To find the opposite of 7+x, find the opposite of each term.

$$2x+5-7-x=6\sqrt{x-2}$$

Subtract 7 from 5 to get -2.

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

Combine 2x and -x to get x.

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

Square both sides of the equation.

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

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

$$x^{2}-4x+4=\left(6\sqrt{x-2}\right)^{2}$$

Expand \left(6\sqrt{x-2}\right)^{2}.

$$x^{2}-4x+4=6^{2}\left(\sqrt{x-2}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

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

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

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

Use the distributive property to multiply 36 by x-2.

$$x^{2}-4x+4=36x-72$$

Subtract 36x from both sides.

$$x^{2}-4x+4-36x=-72$$

Combine -4x and -36x to get -40x.

$$x^{2}-40x+4=-72$$

Add 72 to both sides.

$$x^{2}-40x+4+72=0$$

Add 4 and 72 to get 76.

$$x^{2}-40x+76=0$$

To solve the equation, factor x^{2}-40x+76 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.

$$a+b=-40 ab=76$$

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 76.

$$-1,-76 -2,-38 -4,-19$$

Calculate the sum for each pair.

$$-1-76=-77 -2-38=-40 -4-19=-23$$

The solution is the pair that gives sum -40.

$$a=-38 b=-2$$

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

$$\left(x-38\right)\left(x-2\right)$$

To find equation solutions, solve x-38=0 and x-2=0.

$$x=38 x=2$$

Substitute 38 for x in the equation \sqrt{2x+5}-\sqrt{x-2}=3.

$$\sqrt{2\times 38+5}-\sqrt{38-2}=3$$

Simplify. The value x=38 satisfies the equation.

$$3=3$$

Substitute 2 for x in the equation \sqrt{2x+5}-\sqrt{x-2}=3.

$$\sqrt{2\times 2+5}-\sqrt{2-2}=3$$

Simplify. The value x=2 satisfies the equation.

$$3=3$$

List all solutions of \sqrt{2x+5}=\sqrt{x-2}+3.

$$x=38 x=2$$