Solve sqrt{2x+4}=3+sqrt{x-5}

$$\sqrt { 2 x + 4 } = 3 + \sqrt { x - 5 }$$

Solve for x

$$x=30$$

$$x=6$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x+4}\right)^{2}=\left(3+\sqrt{x-5}\right)^{2}$$

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

$$2x+4=\left(3+\sqrt{x-5}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{x-5}\right)^{2}.

$$2x+4=9+6\sqrt{x-5}+\left(\sqrt{x-5}\right)^{2}$$

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

$$2x+4=9+6\sqrt{x-5}+x-5$$

Subtract 5 from 9 to get 4.

$$2x+4=4+6\sqrt{x-5}+x$$

Subtract 4+x from both sides of the equation.

$$2x+4-\left(4+x\right)=6\sqrt{x-5}$$

To find the opposite of 4+x, find the opposite of each term.

$$2x+4-4-x=6\sqrt{x-5}$$

Subtract 4 from 4 to get 0.

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

Combine 2x and -x to get x.

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

Square both sides of the equation.

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

Expand \left(6\sqrt{x-5}\right)^{2}.

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

Calculate 6 to the power of 2 and get 36.

$$x^{2}=36\left(\sqrt{x-5}\right)^{2}$$

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

$$x^{2}=36\left(x-5\right)$$

Use the distributive property to multiply 36 by x-5.

$$x^{2}=36x-180$$

Subtract 36x from both sides.

$$x^{2}-36x=-180$$

Add 180 to both sides.

$$x^{2}-36x+180=0$$

To solve the equation, factor x^{2}-36x+180 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.

$$a+b=-36 ab=180$$

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 180.

$$-1,-180 -2,-90 -3,-60 -4,-45 -5,-36 -6,-30 -9,-20 -10,-18 -12,-15$$

Calculate the sum for each pair.

$$-1-180=-181 -2-90=-92 -3-60=-63 -4-45=-49 -5-36=-41 -6-30=-36 -9-20=-29 -10-18=-28 -12-15=-27$$

The solution is the pair that gives sum -36.

$$a=-30 b=-6$$

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

$$\left(x-30\right)\left(x-6\right)$$

To find equation solutions, solve x-30=0 and x-6=0.

$$x=30 x=6$$

Substitute 30 for x in the equation \sqrt{2x+4}=3+\sqrt{x-5}.

$$\sqrt{2\times 30+4}=3+\sqrt{30-5}$$

Simplify. The value x=30 satisfies the equation.

$$8=8$$

Substitute 6 for x in the equation \sqrt{2x+4}=3+\sqrt{x-5}.

$$\sqrt{2\times 6+4}=3+\sqrt{6-5}$$

Simplify. The value x=6 satisfies the equation.

$$4=4$$

List all solutions of \sqrt{2x+4}=\sqrt{x-5}+3.

$$x=30 x=6$$