Solve 2-a^2+[1/16(a+1/a)]^2=0

$$2 - a ^ { 2 } + [ \frac { 1 } { 16 } ( a + \frac { 1 } { a } ) ] ^ { 2 } = 0$$

Solve for a (complex solution)

$$a=-\frac{i\sqrt{4080\sqrt{259}-65535}}{255}\approx -0-0.044086859i$$

$$a=\frac{i\sqrt{4080\sqrt{259}-65535}}{255}\approx 0.044086859i$$

$$a = \frac{\sqrt{4080 \sqrt{259} + 65535}}{255} \approx 1.420433006$$

$$a = -\frac{\sqrt{4080 \sqrt{259} + 65535}}{255} \approx -1.420433006$$

Steps Using the Quadratic Formula

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

$$2-a^{2}+\left(\frac{1}{16}\left(\frac{aa}{a}+\frac{1}{a}\right)\right)^{2}=0$$

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

$$2-a^{2}+\left(\frac{1}{16}\times \frac{aa+1}{a}\right)^{2}=0$$

Do the multiplications in aa+1.

$$2-a^{2}+\left(\frac{1}{16}\times \frac{a^{2}+1}{a}\right)^{2}=0$$

Multiply \frac{1}{16} times \frac{a^{2}+1}{a} by multiplying numerator times numerator and denominator times denominator.

$$2-a^{2}+\left(\frac{a^{2}+1}{16a}\right)^{2}=0$$

To raise \frac{a^{2}+1}{16a} to a power, raise both numerator and denominator to the power and then divide.

$$2-a^{2}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2-a^{2} times \frac{\left(16a\right)^{2}}{\left(16a\right)^{2}}.

$$\frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}}{\left(16a\right)^{2}}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Since \frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}}{\left(16a\right)^{2}} and \frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2} times \frac{\left(16a\right)^{2}}{\left(16a\right)^{2}}.

$$2-\frac{a^{2}\times \left(16a\right)^{2}}{\left(16a\right)^{2}}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Since -\frac{a^{2}\times \left(16a\right)^{2}}{\left(16a\right)^{2}} and \frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}} have the same denominator, add them by adding their numerators.

$$2+\frac{-a^{2}\times \left(16a\right)^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Expand \left(16a\right)^{2}.

$$2+\frac{-a^{2}\times 16^{2}a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Calculate 16 to the power of 2 and get 256.

$$2+\frac{-a^{2}\times 256a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Multiply -1 and 256 to get -256.

$$2+\frac{-256a^{2}a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2+\frac{-256a^{4}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

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

$$2+\frac{-256a^{4}+\left(a^{2}\right)^{2}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

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

$$2+\frac{-256a^{4}+a^{4}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

Combine -256a^{4} and a^{4} to get -255a^{4}.

$$2+\frac{-255a^{4}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

Expand \left(16a\right)^{2}.

$$2+\frac{-255a^{4}+2a^{2}+1}{16^{2}a^{2}}=0$$

Calculate 16 to the power of 2 and get 256.

$$2+\frac{-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{256a^{2}}{256a^{2}}.

$$\frac{2\times 256a^{2}}{256a^{2}}+\frac{-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Since \frac{2\times 256a^{2}}{256a^{2}} and \frac{-255a^{4}+2a^{2}+1}{256a^{2}} have the same denominator, add them by adding their numerators.

$$\frac{2\times 256a^{2}-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Do the multiplications in 2\times 256a^{2}-255a^{4}+2a^{2}+1.

$$\frac{512a^{2}-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Combine like terms in 512a^{2}-255a^{4}+2a^{2}+1.

$$\frac{514a^{2}-255a^{4}+1}{256a^{2}}=0$$

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 256a^{2}.

$$514a^{2}-255a^{4}+1=0$$

Substitute t for a^{2}.

$$-255t^{2}+514t+1=0$$

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 -255 for a, 514 for b, and 1 for c in the quadratic formula.

$$t=\frac{-514±\sqrt{514^{2}-4\left(-255\right)\times 1}}{-255\times 2}$$

Do the calculations.

$$t=\frac{-514±32\sqrt{259}}{-510}$$

Solve the equation t=\frac{-514±32\sqrt{259}}{-510} when ± is plus and when ± is minus.

$$t=\frac{257-16\sqrt{259}}{255} t=\frac{16\sqrt{259}+257}{255}$$

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

$$a=-i\sqrt{-\frac{257-16\sqrt{259}}{255}} a=i\sqrt{-\frac{257-16\sqrt{259}}{255}} a=-\sqrt{\frac{16\sqrt{259}+257}{255}} a=\sqrt{\frac{16\sqrt{259}+257}{255}}$$

Solve for a

$$a = \frac{\sqrt{4080 \sqrt{259} + 65535}}{255} \approx 1.420433006$$

$$a = -\frac{\sqrt{4080 \sqrt{259} + 65535}}{255} \approx -1.420433006$$

Steps Using the Quadratic Formula

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

$$2-a^{2}+\left(\frac{1}{16}\left(\frac{aa}{a}+\frac{1}{a}\right)\right)^{2}=0$$

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

$$2-a^{2}+\left(\frac{1}{16}\times \frac{aa+1}{a}\right)^{2}=0$$

Do the multiplications in aa+1.

$$2-a^{2}+\left(\frac{1}{16}\times \frac{a^{2}+1}{a}\right)^{2}=0$$

Multiply \frac{1}{16} times \frac{a^{2}+1}{a} by multiplying numerator times numerator and denominator times denominator.

$$2-a^{2}+\left(\frac{a^{2}+1}{16a}\right)^{2}=0$$

To raise \frac{a^{2}+1}{16a} to a power, raise both numerator and denominator to the power and then divide.

$$2-a^{2}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2-a^{2} times \frac{\left(16a\right)^{2}}{\left(16a\right)^{2}}.

$$\frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}}{\left(16a\right)^{2}}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Since \frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}}{\left(16a\right)^{2}} and \frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(2-a^{2}\right)\times \left(16a\right)^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2} times \frac{\left(16a\right)^{2}}{\left(16a\right)^{2}}.

$$2-\frac{a^{2}\times \left(16a\right)^{2}}{\left(16a\right)^{2}}+\frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Since -\frac{a^{2}\times \left(16a\right)^{2}}{\left(16a\right)^{2}} and \frac{\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}} have the same denominator, add them by adding their numerators.

$$2+\frac{-a^{2}\times \left(16a\right)^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Expand \left(16a\right)^{2}.

$$2+\frac{-a^{2}\times 16^{2}a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Calculate 16 to the power of 2 and get 256.

$$2+\frac{-a^{2}\times 256a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

Multiply -1 and 256 to get -256.

$$2+\frac{-256a^{2}a^{2}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

To multiply powers of the same base, add their exponents. Add 2 and 2 to get 4.

$$2+\frac{-256a^{4}+\left(a^{2}+1\right)^{2}}{\left(16a\right)^{2}}=0$$

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

$$2+\frac{-256a^{4}+\left(a^{2}\right)^{2}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

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

$$2+\frac{-256a^{4}+a^{4}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

Combine -256a^{4} and a^{4} to get -255a^{4}.

$$2+\frac{-255a^{4}+2a^{2}+1}{\left(16a\right)^{2}}=0$$

Expand \left(16a\right)^{2}.

$$2+\frac{-255a^{4}+2a^{2}+1}{16^{2}a^{2}}=0$$

Calculate 16 to the power of 2 and get 256.

$$2+\frac{-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 2 times \frac{256a^{2}}{256a^{2}}.

$$\frac{2\times 256a^{2}}{256a^{2}}+\frac{-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Since \frac{2\times 256a^{2}}{256a^{2}} and \frac{-255a^{4}+2a^{2}+1}{256a^{2}} have the same denominator, add them by adding their numerators.

$$\frac{2\times 256a^{2}-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Do the multiplications in 2\times 256a^{2}-255a^{4}+2a^{2}+1.

$$\frac{512a^{2}-255a^{4}+2a^{2}+1}{256a^{2}}=0$$

Combine like terms in 512a^{2}-255a^{4}+2a^{2}+1.

$$\frac{514a^{2}-255a^{4}+1}{256a^{2}}=0$$

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 256a^{2}.

$$514a^{2}-255a^{4}+1=0$$

Substitute t for a^{2}.

$$-255t^{2}+514t+1=0$$

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 -255 for a, 514 for b, and 1 for c in the quadratic formula.

$$t=\frac{-514±\sqrt{514^{2}-4\left(-255\right)\times 1}}{-255\times 2}$$

Do the calculations.

$$t=\frac{-514±32\sqrt{259}}{-510}$$

Solve the equation t=\frac{-514±32\sqrt{259}}{-510} when ± is plus and when ± is minus.

$$t=\frac{257-16\sqrt{259}}{255} t=\frac{16\sqrt{259}+257}{255}$$

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

$$a=\sqrt{\frac{16\sqrt{259}+257}{255}} a=-\sqrt{\frac{16\sqrt{259}+257}{255}}$$