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

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

Solve for x

$$x = \frac{\sqrt{17} + 9}{8} \approx 1.640388203$$

Solution Steps

Subtract 3 from both sides of the equation.

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

Subtract 3 from 1 to get -2.

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

Square both sides of the equation.

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

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

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

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

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

Subtract 4x^{2} from both sides.

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

Add 8x to both sides.

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

Combine x and 8x to get 9x.

$$9x-4x^{2}=4$$

Subtract 4 from both sides.

$$9x-4x^{2}-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-4x^{2}+9x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 9 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-4\right)\left(-4\right)}}{2\left(-4\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-4\right)\left(-4\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-9±\sqrt{81+16\left(-4\right)}}{2\left(-4\right)}$$

Multiply 16 times -4.

$$x=\frac{-9±\sqrt{81-64}}{2\left(-4\right)}$$

Add 81 to -64.

$$x=\frac{-9±\sqrt{17}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-9±\sqrt{17}}{-8}$$

Now solve the equation x=\frac{-9±\sqrt{17}}{-8} when ± is plus. Add -9 to \sqrt{17}.

$$x=\frac{\sqrt{17}-9}{-8}$$

Divide -9+\sqrt{17} by -8.

$$x=\frac{9-\sqrt{17}}{8}$$

Now solve the equation x=\frac{-9±\sqrt{17}}{-8} when ± is minus. Subtract \sqrt{17} from -9.

$$x=\frac{-\sqrt{17}-9}{-8}$$

Divide -9-\sqrt{17} by -8.

$$x=\frac{\sqrt{17}+9}{8}$$

The equation is now solved.

$$x=\frac{9-\sqrt{17}}{8} x=\frac{\sqrt{17}+9}{8}$$

Substitute \frac{9-\sqrt{17}}{8} for x in the equation \sqrt{x}+3=2x+1.

$$\sqrt{\frac{9-\sqrt{17}}{8}}+3=2\times \frac{9-\sqrt{17}}{8}+1$$

Simplify. The value x=\frac{9-\sqrt{17}}{8} does not satisfy the equation.

$$\frac{11}{4}+\frac{1}{4}\times 17^{\frac{1}{2}}=\frac{13}{4}-\frac{1}{4}\times 17^{\frac{1}{2}}$$

Substitute \frac{\sqrt{17}+9}{8} for x in the equation \sqrt{x}+3=2x+1.

$$\sqrt{\frac{\sqrt{17}+9}{8}}+3=2\times \frac{\sqrt{17}+9}{8}+1$$

Simplify. The value x=\frac{\sqrt{17}+9}{8} satisfies the equation.

$$\frac{13}{4}+\frac{1}{4}\times 17^{\frac{1}{2}}=\frac{1}{4}\times 17^{\frac{1}{2}}+\frac{13}{4}$$

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

$$x=\frac{\sqrt{17}+9}{8}$$