Solve (a-1/a)^2=(2/3)^2

$$( a - \frac { 1 } { a } ) ^ { 2 } = ( \frac { 2 } { 3 } ) ^ { 2 }$$

Solve for a

$$a=\frac{1-\sqrt{10}}{3}\approx -0.72075922$$

$$a = \frac{\sqrt{10} + 1}{3} \approx 1.387425887$$

$$a=\frac{\sqrt{10}-1}{3}\approx 0.72075922$$

$$a=\frac{-\sqrt{10}-1}{3}\approx -1.387425887$$

Steps Using the Quadratic Formula

To add or subtract expressions, expand them to make their denominators the same. Multiply a times \frac{a}{a}.

$$\left(\frac{aa}{a}-\frac{1}{a}\right)^{2}=\left(\frac{2}{3}\right)^{2}$$

Since \frac{aa}{a} and \frac{1}{a} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{aa-1}{a}\right)^{2}=\left(\frac{2}{3}\right)^{2}$$

Do the multiplications in aa-1.

$$\left(\frac{a^{2}-1}{a}\right)^{2}=\left(\frac{2}{3}\right)^{2}$$

To raise \frac{a^{2}-1}{a} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(a^{2}-1\right)^{2}}{a^{2}}=\left(\frac{2}{3}\right)^{2}$$

Calculate \frac{2}{3} to the power of 2 and get \frac{4}{9}.

$$\frac{\left(a^{2}-1\right)^{2}}{a^{2}}=\frac{4}{9}$$

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

$$\frac{\left(a^{2}\right)^{2}-2a^{2}+1}{a^{2}}=\frac{4}{9}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$\frac{a^{4}-2a^{2}+1}{a^{2}}=\frac{4}{9}$$

Subtract \frac{4}{9} from both sides.

$$\frac{a^{4}-2a^{2}+1}{a^{2}}-\frac{4}{9}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a^{2} and 9 is 9a^{2}. Multiply \frac{a^{4}-2a^{2}+1}{a^{2}} times \frac{9}{9}. Multiply \frac{4}{9} times \frac{a^{2}}{a^{2}}.

$$\frac{9\left(a^{4}-2a^{2}+1\right)}{9a^{2}}-\frac{4a^{2}}{9a^{2}}=0$$

Since \frac{9\left(a^{4}-2a^{2}+1\right)}{9a^{2}} and \frac{4a^{2}}{9a^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9\left(a^{4}-2a^{2}+1\right)-4a^{2}}{9a^{2}}=0$$

Do the multiplications in 9\left(a^{4}-2a^{2}+1\right)-4a^{2}.

$$\frac{9a^{4}-18a^{2}+9-4a^{2}}{9a^{2}}=0$$

Combine like terms in 9a^{4}-18a^{2}+9-4a^{2}.

$$\frac{9a^{4}-22a^{2}+9}{9a^{2}}=0$$

Factor the expressions that are not already factored in \frac{9a^{4}-22a^{2}+9}{9a^{2}}.

$$\frac{9\left(a^{2}-\left(-\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)\left(a^{2}-\left(\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)}{9a^{2}}=0$$

Cancel out 9 in both numerator and denominator.

$$\frac{\left(a^{2}-\left(-\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)\left(a^{2}-\left(\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)}{a^{2}}=0$$

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by a^{2}.

$$\left(a^{2}-\left(-\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)\left(a^{2}-\left(\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)=0$$

To find the opposite of -\frac{2}{9}\sqrt{10}+\frac{11}{9}, find the opposite of each term.

$$\left(a^{2}+\frac{2}{9}\sqrt{10}-\frac{11}{9}\right)\left(a^{2}-\left(\frac{2}{9}\sqrt{10}+\frac{11}{9}\right)\right)=0$$

To find the opposite of \frac{2}{9}\sqrt{10}+\frac{11}{9}, find the opposite of each term.

$$\left(a^{2}+\frac{2}{9}\sqrt{10}-\frac{11}{9}\right)\left(a^{2}-\frac{2}{9}\sqrt{10}-\frac{11}{9}\right)=0$$

Use the distributive property to multiply a^{2}+\frac{2}{9}\sqrt{10}-\frac{11}{9} by a^{2}-\frac{2}{9}\sqrt{10}-\frac{11}{9} and combine like terms.

$$a^{4}-\frac{22}{9}a^{2}-\frac{4}{81}\left(\sqrt{10}\right)^{2}+\frac{121}{81}=0$$

The square of \sqrt{10} is 10.

$$a^{4}-\frac{22}{9}a^{2}-\frac{4}{81}\times 10+\frac{121}{81}=0$$

Multiply -\frac{4}{81} and 10 to get -\frac{40}{81}.

$$a^{4}-\frac{22}{9}a^{2}-\frac{40}{81}+\frac{121}{81}=0$$

Add -\frac{40}{81} and \frac{121}{81} to get 1.

$$a^{4}-\frac{22}{9}a^{2}+1=0$$

Substitute t for a^{2}.

$$t^{2}-\frac{22}{9}t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -\frac{22}{9} for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-\frac{22}{9}\right)±\sqrt{\left(-\frac{22}{9}\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{\frac{22}{9}±\frac{4}{9}\sqrt{10}}{2}$$

Solve the equation t=\frac{\frac{22}{9}±\frac{4}{9}\sqrt{10}}{2} when ± is plus and when ± is minus.

$$t=\frac{2\sqrt{10}+11}{9} t=\frac{11-2\sqrt{10}}{9}$$

Since a=t^{2}, the solutions are obtained by evaluating a=±\sqrt{t} for each t.

$$a=\frac{\sqrt{10}+1}{3} a=-\frac{\sqrt{10}+1}{3} a=-\frac{1-\sqrt{10}}{3} a=\frac{1-\sqrt{10}}{3}$$