Solve sqrt{2x+9}-sqrt{x-4}-3=0

$$\sqrt { 2 x + 9 } - \sqrt { x - 4 } - 3 = 0$$

Solve for x

$$x=20$$

$$x=8$$

Solution Steps

Subtract -\sqrt{x-4}-3 from both sides of the equation.

$$\sqrt{2x+9}=-\left(-\sqrt{x-4}-3\right)$$

To find the opposite of -\sqrt{x-4}-3, find the opposite of each term.

$$\sqrt{2x+9}=-\left(-\sqrt{x-4}\right)-\left(-3\right)$$

The opposite of -\sqrt{x-4} is \sqrt{x-4}.

$$\sqrt{2x+9}=\sqrt{x-4}-\left(-3\right)$$

The opposite of -3 is 3.

$$\sqrt{2x+9}=\sqrt{x-4}+3$$

Square both sides of the equation.

$$\left(\sqrt{2x+9}\right)^{2}=\left(\sqrt{x-4}+3\right)^{2}$$

Calculate \sqrt{2x+9} to the power of 2 and get 2x+9.

$$2x+9=\left(\sqrt{x-4}+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x-4}+3\right)^{2}.

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

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

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

Add -4 and 9 to get 5.

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

Subtract x+5 from both sides of the equation.

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

To find the opposite of x+5, find the opposite of each term.

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

Combine 2x and -x to get x.

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

Subtract 5 from 9 to get 4.

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

Square both sides of the equation.

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

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

$$x^{2}+8x+16=\left(6\sqrt{x-4}\right)^{2}$$

Expand \left(6\sqrt{x-4}\right)^{2}.

$$x^{2}+8x+16=6^{2}\left(\sqrt{x-4}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$x^{2}+8x+16=36\left(\sqrt{x-4}\right)^{2}$$

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

$$x^{2}+8x+16=36\left(x-4\right)$$

Use the distributive property to multiply 36 by x-4.

$$x^{2}+8x+16=36x-144$$

Subtract 36x from both sides.

$$x^{2}+8x+16-36x=-144$$

Combine 8x and -36x to get -28x.

$$x^{2}-28x+16=-144$$

Add 144 to both sides.

$$x^{2}-28x+16+144=0$$

Add 16 and 144 to get 160.

$$x^{2}-28x+160=0$$

To solve the equation, factor x^{2}-28x+160 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.

$$a+b=-28 ab=160$$

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 160.

$$-1,-160 -2,-80 -4,-40 -5,-32 -8,-20 -10,-16$$

Calculate the sum for each pair.

$$-1-160=-161 -2-80=-82 -4-40=-44 -5-32=-37 -8-20=-28 -10-16=-26$$

The solution is the pair that gives sum -28.

$$a=-20 b=-8$$

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

$$\left(x-20\right)\left(x-8\right)$$

To find equation solutions, solve x-20=0 and x-8=0.

$$x=20 x=8$$

Substitute 20 for x in the equation \sqrt{2x+9}-\sqrt{x-4}-3=0.

$$\sqrt{2\times 20+9}-\sqrt{20-4}-3=0$$

Simplify. The value x=20 satisfies the equation.

$$0=0$$

Substitute 8 for x in the equation \sqrt{2x+9}-\sqrt{x-4}-3=0.

$$\sqrt{2\times 8+9}-\sqrt{8-4}-3=0$$

Simplify. The value x=8 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{2x+9}=\sqrt{x-4}+3.

$$x=20 x=8$$