Solve {left(3x-1/3right)}^2-sqrt{5/18+5/2}=-10/6

$${ \left(3x- \frac{ 1 }{ 3 } \right) }^{ 2 } - \sqrt{ \frac{ 5 }{ 18 } + \frac{ 5 }{ 2 } } = - \frac{ 10 }{ 6 }$$

Solve for x

$$x=\frac{1}{9}\approx 0.111111111$$

Steps Using the Quadratic Formula

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

$$9x^{2}-2x+\frac{1}{9}-\sqrt{\frac{5}{18}+\frac{5}{2}}=-\frac{10}{6}$$

Add \frac{5}{18} and \frac{5}{2} to get \frac{25}{9}.

$$9x^{2}-2x+\frac{1}{9}-\sqrt{\frac{25}{9}}=-\frac{10}{6}$$

Rewrite the square root of the division \frac{25}{9} as the division of square roots \frac{\sqrt{25}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$9x^{2}-2x+\frac{1}{9}-\frac{5}{3}=-\frac{10}{6}$$

Subtract \frac{5}{3} from \frac{1}{9} to get -\frac{14}{9}.

$$9x^{2}-2x-\frac{14}{9}=-\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$9x^{2}-2x-\frac{14}{9}=-\frac{5}{3}$$

Add \frac{5}{3} to both sides.

$$9x^{2}-2x-\frac{14}{9}+\frac{5}{3}=0$$

Add -\frac{14}{9} and \frac{5}{3} to get \frac{1}{9}.

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

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

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

Square -2.

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

Multiply -4 times 9.

$$x=\frac{-\left(-2\right)±\sqrt{4-36\times \frac{1}{9}}}{2\times 9}$$

Multiply -36 times \frac{1}{9}.

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

Add 4 to -4.

$$x=\frac{-\left(-2\right)±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$x=-\frac{-2}{2\times 9}$$

The opposite of -2 is 2.

$$x=\frac{2}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{2}{18}$$

Reduce the fraction \frac{2}{18} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{9}$$

Steps for Completing the Square

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

$$9x^{2}-2x+\frac{1}{9}-\sqrt{\frac{5}{18}+\frac{5}{2}}=-\frac{10}{6}$$

Add \frac{5}{18} and \frac{5}{2} to get \frac{25}{9}.

$$9x^{2}-2x+\frac{1}{9}-\sqrt{\frac{25}{9}}=-\frac{10}{6}$$

Rewrite the square root of the division \frac{25}{9} as the division of square roots \frac{\sqrt{25}}{\sqrt{9}}. Take the square root of both numerator and denominator.

$$9x^{2}-2x+\frac{1}{9}-\frac{5}{3}=-\frac{10}{6}$$

Subtract \frac{5}{3} from \frac{1}{9} to get -\frac{14}{9}.

$$9x^{2}-2x-\frac{14}{9}=-\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

$$9x^{2}-2x-\frac{14}{9}=-\frac{5}{3}$$

Add \frac{14}{9} to both sides.

$$9x^{2}-2x=-\frac{5}{3}+\frac{14}{9}$$

Add -\frac{5}{3} and \frac{14}{9} to get -\frac{1}{9}.

$$9x^{2}-2x=-\frac{1}{9}$$

Divide both sides by 9.

$$\frac{9x^{2}-2x}{9}=-\frac{\frac{1}{9}}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{2}{9}x=-\frac{\frac{1}{9}}{9}$$

Divide -\frac{1}{9} by 9.

$$x^{2}-\frac{2}{9}x=-\frac{1}{81}$$

Divide -\frac{2}{9}, the coefficient of the x term, by 2 to get -\frac{1}{9}. Then add the square of -\frac{1}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{9}x+\left(-\frac{1}{9}\right)^{2}=-\frac{1}{81}+\left(-\frac{1}{9}\right)^{2}$$

Square -\frac{1}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{9}x+\frac{1}{81}=\frac{-1+1}{81}$$

Add -\frac{1}{81} to \frac{1}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{9}x+\frac{1}{81}=0$$

Factor x^{2}-\frac{2}{9}x+\frac{1}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{9}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{9}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{9}=0 x-\frac{1}{9}=0$$

Add \frac{1}{9} to both sides of the equation.

$$x=\frac{1}{9} x=\frac{1}{9}$$

The equation is now solved. Solutions are the same.

$$x=\frac{1}{9}$$