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

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

Solve for x

$$x=4$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Combine 2x and x to get 3x.

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

Subtract 3 from 1 to get -2.

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

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

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

Subtract 3x-2 from both sides of the equation.

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

To find the opposite of 3x-2, find the opposite of each term.

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

Combine 4x and -3x to get x.

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

Square both sides of the equation.

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

Expand \left(2\sqrt{2x+1}\sqrt{x-3}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

Use the distributive property to multiply 4 by 2x+1.

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

Apply the distributive property by multiplying each term of 8x+4 by each term of x-3.

$$8x^{2}-24x+4x-12=\left(x+2\right)^{2}$$

Combine -24x and 4x to get -20x.

$$8x^{2}-20x-12=\left(x+2\right)^{2}$$

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

$$8x^{2}-20x-12=x^{2}+4x+4$$

Subtract x^{2} from both sides.

$$8x^{2}-20x-12-x^{2}=4x+4$$

Combine 8x^{2} and -x^{2} to get 7x^{2}.

$$7x^{2}-20x-12=4x+4$$

Subtract 4x from both sides.

$$7x^{2}-20x-12-4x=4$$

Combine -20x and -4x to get -24x.

$$7x^{2}-24x-12=4$$

Subtract 4 from both sides.

$$7x^{2}-24x-12-4=0$$

Subtract 4 from -12 to get -16.

$$7x^{2}-24x-16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 7x^{2}+ax+bx-16. To find a and b, set up a system to be solved.

$$a+b=-24 ab=7\left(-16\right)=-112$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -112.

$$1,-112 2,-56 4,-28 7,-16 8,-14$$

Calculate the sum for each pair.

$$1-112=-111 2-56=-54 4-28=-24 7-16=-9 8-14=-6$$

The solution is the pair that gives sum -24.

$$a=-28 b=4$$

Rewrite 7x^{2}-24x-16 as \left(7x^{2}-28x\right)+\left(4x-16\right).

$$\left(7x^{2}-28x\right)+\left(4x-16\right)$$

Factor out 7x in the first and 4 in the second group.

$$7x\left(x-4\right)+4\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(7x+4\right)$$

To find equation solutions, solve x-4=0 and 7x+4=0.

$$x=4 x=-\frac{4}{7}$$

Substitute -\frac{4}{7} for x in the equation \sqrt{2x+1}+\sqrt{x-3}=\sqrt{4x}. The expression \sqrt{2\left(-\frac{4}{7}\right)+1} is undefined because the radicand cannot be negative.

$$\sqrt{2\left(-\frac{4}{7}\right)+1}+\sqrt{-\frac{4}{7}-3}=\sqrt{4\left(-\frac{4}{7}\right)}$$

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

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

Simplify. The value x=4 satisfies the equation.

$$4=4$$

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

$$x=4$$