Solve sqrt{3x}-sqrt{x}=2

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

Solve for x

$$x=2\left(\sqrt{3}+2\right)\approx 7.464101615$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 4+x from both sides of the equation.

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

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

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

Combine 3x and -x to get 2x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 4 to the power of 2 and get 16.

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

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

$$4x^{2}-16x+16=16x$$

Subtract 16x from both sides.

$$4x^{2}-16x+16-16x=0$$

Combine -16x and -16x to get -32x.

$$4x^{2}-32x+16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -32 for b, and 16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 4\times 16}}{2\times 4}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 4\times 16}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-32\right)±\sqrt{1024-16\times 16}}{2\times 4}$$

Multiply -16 times 16.

$$x=\frac{-\left(-32\right)±\sqrt{1024-256}}{2\times 4}$$

Add 1024 to -256.

$$x=\frac{-\left(-32\right)±\sqrt{768}}{2\times 4}$$

Take the square root of 768.

$$x=\frac{-\left(-32\right)±16\sqrt{3}}{2\times 4}$$

The opposite of -32 is 32.

$$x=\frac{32±16\sqrt{3}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{32±16\sqrt{3}}{8}$$

Now solve the equation x=\frac{32±16\sqrt{3}}{8} when ± is plus. Add 32 to 16\sqrt{3}.

$$x=\frac{16\sqrt{3}+32}{8}$$

Divide 32+16\sqrt{3} by 8.

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

Now solve the equation x=\frac{32±16\sqrt{3}}{8} when ± is minus. Subtract 16\sqrt{3} from 32.

$$x=\frac{32-16\sqrt{3}}{8}$$

Divide 32-16\sqrt{3} by 8.

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

The equation is now solved.

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

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

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

Simplify. The value x=2\sqrt{3}+4 satisfies the equation.

$$2=2$$

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

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

Simplify. The value x=4-2\sqrt{3} does not satisfy the equation.

$$4-2\times 3^{\frac{1}{2}}=2$$

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

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

Simplify. The value x=2\sqrt{3}+4 satisfies the equation.

$$2=2$$

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

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