Solve sqrt{x}+2=5x

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

Solve for x

$$x=\frac{\sqrt{41}+21}{50}\approx 0.548062485$$

Solution Steps

Subtract 2 from both sides of the equation.

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

Square both sides of the equation.

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

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

$$x=\left(5x-2\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-2\right)^{2}.

$$x=25x^{2}-20x+4$$

Subtract 25x^{2} from both sides.

$$x-25x^{2}=-20x+4$$

Add 20x to both sides.

$$x-25x^{2}+20x=4$$

Combine x and 20x to get 21x.

$$21x-25x^{2}=4$$

Subtract 4 from both sides.

$$21x-25x^{2}-4=0$$

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.

$$-25x^{2}+21x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 21 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-21±\sqrt{21^{2}-4\left(-25\right)\left(-4\right)}}{2\left(-25\right)}$$

Square 21.

$$x=\frac{-21±\sqrt{441-4\left(-25\right)\left(-4\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-21±\sqrt{441+100\left(-4\right)}}{2\left(-25\right)}$$

Multiply 100 times -4.

$$x=\frac{-21±\sqrt{441-400}}{2\left(-25\right)}$$

Add 441 to -400.

$$x=\frac{-21±\sqrt{41}}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-21±\sqrt{41}}{-50}$$

Now solve the equation x=\frac{-21±\sqrt{41}}{-50} when ± is plus. Add -21 to \sqrt{41}.

$$x=\frac{\sqrt{41}-21}{-50}$$

Divide -21+\sqrt{41} by -50.

$$x=\frac{21-\sqrt{41}}{50}$$

Now solve the equation x=\frac{-21±\sqrt{41}}{-50} when ± is minus. Subtract \sqrt{41} from -21.

$$x=\frac{-\sqrt{41}-21}{-50}$$

Divide -21-\sqrt{41} by -50.

$$x=\frac{\sqrt{41}+21}{50}$$

The equation is now solved.

$$x=\frac{21-\sqrt{41}}{50} x=\frac{\sqrt{41}+21}{50}$$

Substitute \frac{21-\sqrt{41}}{50} for x in the equation \sqrt{x}+2=5x.

$$\sqrt{\frac{21-\sqrt{41}}{50}}+2=5\times \frac{21-\sqrt{41}}{50}$$

Simplify. The value x=\frac{21-\sqrt{41}}{50} does not satisfy the equation.

$$\frac{19}{10}+\frac{1}{10}\times 41^{\frac{1}{2}}=\frac{21}{10}-\frac{1}{10}\times 41^{\frac{1}{2}}$$

Substitute \frac{\sqrt{41}+21}{50} for x in the equation \sqrt{x}+2=5x.

$$\sqrt{\frac{\sqrt{41}+21}{50}}+2=5\times \frac{\sqrt{41}+21}{50}$$

Simplify. The value x=\frac{\sqrt{41}+21}{50} satisfies the equation.

$$\frac{21}{10}+\frac{1}{10}\times 41^{\frac{1}{2}}=\frac{1}{10}\times 41^{\frac{1}{2}}+\frac{21}{10}$$

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

$$x=\frac{\sqrt{41}+21}{50}$$