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

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

Solve for x

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

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 1+x from both sides of the equation.

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

To find the opposite of 1+x, find the opposite of each term.

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

Subtract 1 from -1 to get -2.

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

Combine 2x and -x to get x.

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

Square both sides of the equation.

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Subtract 4x from both sides.

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

Combine -4x and -4x to get -8x.

$$x^{2}-8x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 4}}{2}$$

Multiply -4 times 4.

$$x=\frac{-\left(-8\right)±\sqrt{64-16}}{2}$$

Add 64 to -16.

$$x=\frac{-\left(-8\right)±\sqrt{48}}{2}$$

Take the square root of 48.

$$x=\frac{-\left(-8\right)±4\sqrt{3}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{3}}{2}$$

Now solve the equation x=\frac{8±4\sqrt{3}}{2} when ± is plus. Add 8 to 4\sqrt{3}.

$$x=\frac{4\sqrt{3}+8}{2}$$

Divide 8+4\sqrt{3} by 2.

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

Now solve the equation x=\frac{8±4\sqrt{3}}{2} when ± is minus. Subtract 4\sqrt{3} from 8.

$$x=\frac{8-4\sqrt{3}}{2}$$

Divide 8-4\sqrt{3} by 2.

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

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{2x-1}=1+\sqrt{x}.

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

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

$$2+3^{\frac{1}{2}}=2+3^{\frac{1}{2}}$$

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

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

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

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

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

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

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

$$2+3^{\frac{1}{2}}=2+3^{\frac{1}{2}}$$

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

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