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

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

Solve for x

$$x=26-6\sqrt{17}\approx 1.261366246$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

Subtract 9+x from both sides of the equation.

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

To find the opposite of 9+x, find the opposite of each term.

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

Subtract 9 from 1 to get -8.

$$2x-8-x=-6\sqrt{x}$$

Combine 2x and -x to get x.

$$x-8=-6\sqrt{x}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-8\right)^{2}.

$$x^{2}-16x+64=\left(-6\sqrt{x}\right)^{2}$$

Expand \left(-6\sqrt{x}\right)^{2}.

$$x^{2}-16x+64=\left(-6\right)^{2}\left(\sqrt{x}\right)^{2}$$

Calculate -6 to the power of 2 and get 36.

$$x^{2}-16x+64=36\left(\sqrt{x}\right)^{2}$$

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

$$x^{2}-16x+64=36x$$

Subtract 36x from both sides.

$$x^{2}-16x+64-36x=0$$

Combine -16x and -36x to get -52x.

$$x^{2}-52x+64=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -52 for b, and 64 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-52\right)±\sqrt{\left(-52\right)^{2}-4\times 64}}{2}$$

Square -52.

$$x=\frac{-\left(-52\right)±\sqrt{2704-4\times 64}}{2}$$

Multiply -4 times 64.

$$x=\frac{-\left(-52\right)±\sqrt{2704-256}}{2}$$

Add 2704 to -256.

$$x=\frac{-\left(-52\right)±\sqrt{2448}}{2}$$

Take the square root of 2448.

$$x=\frac{-\left(-52\right)±12\sqrt{17}}{2}$$

The opposite of -52 is 52.

$$x=\frac{52±12\sqrt{17}}{2}$$

Now solve the equation x=\frac{52±12\sqrt{17}}{2} when ± is plus. Add 52 to 12\sqrt{17}.

$$x=\frac{12\sqrt{17}+52}{2}$$

Divide 52+12\sqrt{17} by 2.

$$x=6\sqrt{17}+26$$

Now solve the equation x=\frac{52±12\sqrt{17}}{2} when ± is minus. Subtract 12\sqrt{17} from 52.

$$x=\frac{52-12\sqrt{17}}{2}$$

Divide 52-12\sqrt{17} by 2.

$$x=26-6\sqrt{17}$$

The equation is now solved.

$$x=6\sqrt{17}+26 x=26-6\sqrt{17}$$

Substitute 6\sqrt{17}+26 for x in the equation \sqrt{2x+1}+\sqrt{x}=3.

$$\sqrt{2\left(6\sqrt{17}+26\right)+1}+\sqrt{6\sqrt{17}+26}=3$$

Simplify. The value x=6\sqrt{17}+26 does not satisfy the equation.

$$9+2\times 17^{\frac{1}{2}}=3$$

Substitute 26-6\sqrt{17} for x in the equation \sqrt{2x+1}+\sqrt{x}=3.

$$\sqrt{2\left(26-6\sqrt{17}\right)+1}+\sqrt{26-6\sqrt{17}}=3$$

Simplify. The value x=26-6\sqrt{17} satisfies the equation.

$$3=3$$

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

$$x=26-6\sqrt{17}$$