Solve sqrt{5x+1}-sqrt{x}=2

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

Solve for x

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

Solution Steps

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

$$\sqrt{5x+1}=2+\sqrt{x}$$

Square both sides of the equation.

$$\left(\sqrt{5x+1}\right)^{2}=\left(2+\sqrt{x}\right)^{2}$$

Calculate \sqrt{5x+1} to the power of 2 and get 5x+1.

$$5x+1=\left(2+\sqrt{x}\right)^{2}$$

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

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

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

$$5x+1=4+4\sqrt{x}+x$$

Subtract 4+x from both sides of the equation.

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

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

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

Subtract 4 from 1 to get -3.

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

Combine 5x and -x to get 4x.

$$4x-3=4\sqrt{x}$$

Square both sides of the equation.

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

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

$$16x^{2}-24x+9=\left(4\sqrt{x}\right)^{2}$$

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

$$16x^{2}-24x+9=4^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16x^{2}-24x+9=16\left(\sqrt{x}\right)^{2}$$

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

$$16x^{2}-24x+9=16x$$

Subtract 16x from both sides.

$$16x^{2}-24x+9-16x=0$$

Combine -24x and -16x to get -40x.

$$16x^{2}-40x+9=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx+9. To find a and b, set up a system to be solved.

$$a+b=-40 ab=16\times 9=144$$

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 144.

$$-1,-144 -2,-72 -3,-48 -4,-36 -6,-24 -8,-18 -9,-16 -12,-12$$

Calculate the sum for each pair.

$$-1-144=-145 -2-72=-74 -3-48=-51 -4-36=-40 -6-24=-30 -8-18=-26 -9-16=-25 -12-12=-24$$

The solution is the pair that gives sum -40.

$$a=-36 b=-4$$

Rewrite 16x^{2}-40x+9 as \left(16x^{2}-36x\right)+\left(-4x+9\right).

$$\left(16x^{2}-36x\right)+\left(-4x+9\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(4x-9\right)-\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(4x-1\right)$$

To find equation solutions, solve 4x-9=0 and 4x-1=0.

$$x=\frac{9}{4} x=\frac{1}{4}$$

Substitute \frac{9}{4} for x in the equation \sqrt{5x+1}-\sqrt{x}=2.

$$\sqrt{5\times \frac{9}{4}+1}-\sqrt{\frac{9}{4}}=2$$

Simplify. The value x=\frac{9}{4} satisfies the equation.

$$2=2$$

Substitute \frac{1}{4} for x in the equation \sqrt{5x+1}-\sqrt{x}=2.

$$\sqrt{5\times \frac{1}{4}+1}-\sqrt{\frac{1}{4}}=2$$

Simplify. The value x=\frac{1}{4} does not satisfy the equation.

$$1=2$$

Substitute \frac{9}{4} for x in the equation \sqrt{5x+1}-\sqrt{x}=2.

$$\sqrt{5\times \frac{9}{4}+1}-\sqrt{\frac{9}{4}}=2$$

Simplify. The value x=\frac{9}{4} satisfies the equation.

$$2=2$$

Equation \sqrt{5x+1}=\sqrt{x}+2 has a unique solution.

$$x=\frac{9}{4}$$