Solve sqrt{2x+1}+sqrt{x-4}=3

$$\sqrt{ 2x+1 } + \sqrt{ x-4 } =3$$

Solve for x

$$x=4$$

Solution Steps

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

$$\sqrt{2x+1}=3-\sqrt{x-4}$$

Square both sides of the equation.

$$\left(\sqrt{2x+1}\right)^{2}=\left(3-\sqrt{x-4}\right)^{2}$$

Calculate \sqrt{2x+1} to the power of 2 and get 2x+1.

$$2x+1=\left(3-\sqrt{x-4}\right)^{2}$$

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

$$2x+1=9-6\sqrt{x-4}+\left(\sqrt{x-4}\right)^{2}$$

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

$$2x+1=9-6\sqrt{x-4}+x-4$$

Subtract 4 from 9 to get 5.

$$2x+1=5-6\sqrt{x-4}+x$$

Subtract 5+x from both sides of the equation.

$$2x+1-\left(5+x\right)=-6\sqrt{x-4}$$

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

$$2x+1-5-x=-6\sqrt{x-4}$$

Subtract 5 from 1 to get -4.

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

Combine 2x and -x to get x.

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

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}.

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

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 -44x.

$$x^{2}-44x+16=-144$$

Add 144 to both sides.

$$x^{2}-44x+16+144=0$$

Add 16 and 144 to get 160.

$$x^{2}-44x+160=0$$

To solve the equation, factor x^{2}-44x+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.

$$a+b=-44 ab=160$$

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 -44.

$$a=-40 b=-4$$

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

$$\left(x-40\right)\left(x-4\right)$$

To find equation solutions, solve x-40=0 and x-4=0.

$$x=40 x=4$$

Substitute 40 for x in the equation \sqrt{2x+1}+\sqrt{x-4}=3.

$$\sqrt{2\times 40+1}+\sqrt{40-4}=3$$

Simplify. The value x=40 does not satisfy the equation.

$$15=3$$

Substitute 4 for x in the equation \sqrt{2x+1}+\sqrt{x-4}=3.

$$\sqrt{2\times 4+1}+\sqrt{4-4}=3$$

Simplify. The value x=4 satisfies the equation.

$$3=3$$

Equation \sqrt{2x+1}=-\sqrt{x-4}+3 has a unique solution.

$$x=4$$