Solve 3sqrt{x}+1-3x=0

$$3 \sqrt { x } + 1 - 3 x = 0$$

Solve for x

$$x = \frac{\sqrt{21} + 5}{6} \approx 1.597095949$$

Solution Steps

Subtract 1-3x from both sides of the equation.

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

To find the opposite of 1-3x, find the opposite of each term.

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

The opposite of -3x is 3x.

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

Square both sides of the equation.

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

Expand \left(3\sqrt{x}\right)^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$9\left(\sqrt{x}\right)^{2}=\left(-1+3x\right)^{2}$$

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

$$9x=\left(-1+3x\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-1+3x\right)^{2}.

$$9x=1-6x+9x^{2}$$

Subtract 1 from both sides.

$$9x-1=-6x+9x^{2}$$

Add 6x to both sides.

$$9x-1+6x=9x^{2}$$

Combine 9x and 6x to get 15x.

$$15x-1=9x^{2}$$

Subtract 9x^{2} from both sides.

$$15x-1-9x^{2}=0$$

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.

$$-9x^{2}+15x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 15 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-15±\sqrt{15^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Square 15.

$$x=\frac{-15±\sqrt{225-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-15±\sqrt{225+36\left(-1\right)}}{2\left(-9\right)}$$

Multiply 36 times -1.

$$x=\frac{-15±\sqrt{225-36}}{2\left(-9\right)}$$

Add 225 to -36.

$$x=\frac{-15±\sqrt{189}}{2\left(-9\right)}$$

Take the square root of 189.

$$x=\frac{-15±3\sqrt{21}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-15±3\sqrt{21}}{-18}$$

Now solve the equation x=\frac{-15±3\sqrt{21}}{-18} when ± is plus. Add -15 to 3\sqrt{21}.

$$x=\frac{3\sqrt{21}-15}{-18}$$

Divide -15+3\sqrt{21} by -18.

$$x=\frac{5-\sqrt{21}}{6}$$

Now solve the equation x=\frac{-15±3\sqrt{21}}{-18} when ± is minus. Subtract 3\sqrt{21} from -15.

$$x=\frac{-3\sqrt{21}-15}{-18}$$

Divide -15-3\sqrt{21} by -18.

$$x=\frac{\sqrt{21}+5}{6}$$

The equation is now solved.

$$x=\frac{5-\sqrt{21}}{6} x=\frac{\sqrt{21}+5}{6}$$

Substitute \frac{5-\sqrt{21}}{6} for x in the equation 3\sqrt{x}+1-3x=0.

$$3\sqrt{\frac{5-\sqrt{21}}{6}}+1-3\times \frac{5-\sqrt{21}}{6}=0$$

Simplify. The value x=\frac{5-\sqrt{21}}{6} does not satisfy the equation.

$$-3+21^{\frac{1}{2}}=0$$

Substitute \frac{\sqrt{21}+5}{6} for x in the equation 3\sqrt{x}+1-3x=0.

$$3\sqrt{\frac{\sqrt{21}+5}{6}}+1-3\times \frac{\sqrt{21}+5}{6}=0$$

Simplify. The value x=\frac{\sqrt{21}+5}{6} satisfies the equation.

$$0=0$$

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

$$x=\frac{\sqrt{21}+5}{6}$$