Solve (4/6)=2/5(5/6)^2+1/3u^2

$$( \frac { 4 } { 6 } ) = \frac { 2 } { 5 } ( \frac { 5 } { 6 } ) ^ { 2 } + \frac { 1 } { 3 } u ^ { 2 }$$

Solve for u

$$u = \frac{\sqrt{42}}{6} \approx 1.08012345$$

$$u = -\frac{\sqrt{42}}{6} \approx -1.08012345$$

Steps by Finding Square Root

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}=\frac{2}{5}\times \left(\frac{5}{6}\right)^{2}+\frac{1}{3}u^{2}$$

Calculate \frac{5}{6} to the power of 2 and get \frac{25}{36}.

$$\frac{2}{3}=\frac{2}{5}\times \frac{25}{36}+\frac{1}{3}u^{2}$$

Multiply \frac{2}{5} and \frac{25}{36} to get \frac{5}{18}.

$$\frac{2}{3}=\frac{5}{18}+\frac{1}{3}u^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{5}{18}+\frac{1}{3}u^{2}=\frac{2}{3}$$

Subtract \frac{5}{18} from both sides.

$$\frac{1}{3}u^{2}=\frac{2}{3}-\frac{5}{18}$$

Subtract \frac{5}{18} from \frac{2}{3} to get \frac{7}{18}.

$$\frac{1}{3}u^{2}=\frac{7}{18}$$

Multiply both sides by 3, the reciprocal of \frac{1}{3}.

$$u^{2}=\frac{7}{18}\times 3$$

Multiply \frac{7}{18} and 3 to get \frac{7}{6}.

$$u^{2}=\frac{7}{6}$$

Take the square root of both sides of the equation.

$$u=\frac{\sqrt{42}}{6} u=-\frac{\sqrt{42}}{6}$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{4}{6} to lowest terms by extracting and canceling out 2.

$$\frac{2}{3}=\frac{2}{5}\times \left(\frac{5}{6}\right)^{2}+\frac{1}{3}u^{2}$$

Calculate \frac{5}{6} to the power of 2 and get \frac{25}{36}.

$$\frac{2}{3}=\frac{2}{5}\times \frac{25}{36}+\frac{1}{3}u^{2}$$

Multiply \frac{2}{5} and \frac{25}{36} to get \frac{5}{18}.

$$\frac{2}{3}=\frac{5}{18}+\frac{1}{3}u^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{5}{18}+\frac{1}{3}u^{2}=\frac{2}{3}$$

Subtract \frac{2}{3} from both sides.

$$\frac{5}{18}+\frac{1}{3}u^{2}-\frac{2}{3}=0$$

Subtract \frac{2}{3} from \frac{5}{18} to get -\frac{7}{18}.

$$-\frac{7}{18}+\frac{1}{3}u^{2}=0$$

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$\frac{1}{3}u^{2}-\frac{7}{18}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{3} for a, 0 for b, and -\frac{7}{18} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$u=\frac{0±\sqrt{0^{2}-4\times \frac{1}{3}\left(-\frac{7}{18}\right)}}{2\times \frac{1}{3}}$$

Square 0.

$$u=\frac{0±\sqrt{-4\times \frac{1}{3}\left(-\frac{7}{18}\right)}}{2\times \frac{1}{3}}$$

Multiply -4 times \frac{1}{3}.

$$u=\frac{0±\sqrt{-\frac{4}{3}\left(-\frac{7}{18}\right)}}{2\times \frac{1}{3}}$$

Multiply -\frac{4}{3} times -\frac{7}{18} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$u=\frac{0±\sqrt{\frac{14}{27}}}{2\times \frac{1}{3}}$$

Take the square root of \frac{14}{27}.

$$u=\frac{0±\frac{\sqrt{42}}{9}}{2\times \frac{1}{3}}$$

Multiply 2 times \frac{1}{3}.

$$u=\frac{0±\frac{\sqrt{42}}{9}}{\frac{2}{3}}$$

Now solve the equation u=\frac{0±\frac{\sqrt{42}}{9}}{\frac{2}{3}} when ± is plus.

$$u=\frac{\sqrt{42}}{6}$$

Now solve the equation u=\frac{0±\frac{\sqrt{42}}{9}}{\frac{2}{3}} when ± is minus.

$$u=-\frac{\sqrt{42}}{6}$$

The equation is now solved.

$$u=\frac{\sqrt{42}}{6} u=-\frac{\sqrt{42}}{6}$$