Solve a^2+4+80=(2+sqrt{80-a^2})^2

$$a ^ { 2 } + 4 + 80 = ( 2 + \sqrt { 80 - a ^ { 2 } } ) ^ { 2 }$$

Solve for a

$$a=4$$

$$a=-4$$

Solution Steps

Add 4 and 80 to get 84.

$$a^{2}+84=\left(2+\sqrt{80-a^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{80-a^{2}}\right)^{2}.

$$a^{2}+84=4+4\sqrt{80-a^{2}}+\left(\sqrt{80-a^{2}}\right)^{2}$$

Calculate \sqrt{80-a^{2}} to the power of 2 and get 80-a^{2}.

$$a^{2}+84=4+4\sqrt{80-a^{2}}+80-a^{2}$$

Add 4 and 80 to get 84.

$$a^{2}+84=84+4\sqrt{80-a^{2}}-a^{2}$$

Subtract 4\sqrt{80-a^{2}} from both sides.

$$a^{2}+84-4\sqrt{80-a^{2}}=84-a^{2}$$

Add a^{2} to both sides.

$$a^{2}+84-4\sqrt{80-a^{2}}+a^{2}=84$$

Combine a^{2} and a^{2} to get 2a^{2}.

$$2a^{2}+84-4\sqrt{80-a^{2}}=84$$

Subtract 2a^{2}+84 from both sides of the equation.

$$-4\sqrt{80-a^{2}}=84-\left(2a^{2}+84\right)$$

To find the opposite of 2a^{2}+84, find the opposite of each term.

$$-4\sqrt{80-a^{2}}=84-2a^{2}-84$$

Subtract 84 from 84 to get 0.

$$-4\sqrt{80-a^{2}}=-2a^{2}$$

Square both sides of the equation.

$$\left(-4\sqrt{80-a^{2}}\right)^{2}=\left(-2a^{2}\right)^{2}$$

Expand \left(-4\sqrt{80-a^{2}}\right)^{2}.

$$\left(-4\right)^{2}\left(\sqrt{80-a^{2}}\right)^{2}=\left(-2a^{2}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$16\left(\sqrt{80-a^{2}}\right)^{2}=\left(-2a^{2}\right)^{2}$$

Calculate \sqrt{80-a^{2}} to the power of 2 and get 80-a^{2}.

$$16\left(80-a^{2}\right)=\left(-2a^{2}\right)^{2}$$

Use the distributive property to multiply 16 by 80-a^{2}.

$$1280-16a^{2}=\left(-2a^{2}\right)^{2}$$

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

$$1280-16a^{2}=\left(-2\right)^{2}\left(a^{2}\right)^{2}$$

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

$$1280-16a^{2}=\left(-2\right)^{2}a^{4}$$

Calculate -2 to the power of 2 and get 4.

$$1280-16a^{2}=4a^{4}$$

Subtract 4a^{4} from both sides.

$$1280-16a^{2}-4a^{4}=0$$

Substitute t for a^{2}.

$$-4t^{2}-16t+1280=0$$

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 -4 for a, -16 for b, and 1280 for c in the quadratic formula.

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-4\right)\times 1280}}{-4\times 2}$$

Do the calculations.

$$t=\frac{16±144}{-8}$$

Solve the equation t=\frac{16±144}{-8} when ± is plus and when ± is minus.

$$t=-20 t=16$$

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

$$a=4 a=-4$$

Substitute 4 for a in the equation a^{2}+4+80=\left(2+\sqrt{80-a^{2}}\right)^{2}.

$$4^{2}+4+80=\left(2+\sqrt{80-4^{2}}\right)^{2}$$

Simplify. The value a=4 satisfies the equation.

$$100=100$$

Substitute -4 for a in the equation a^{2}+4+80=\left(2+\sqrt{80-a^{2}}\right)^{2}.

$$\left(-4\right)^{2}+4+80=\left(2+\sqrt{80-\left(-4\right)^{2}}\right)^{2}$$

Simplify. The value a=-4 satisfies the equation.

$$100=100$$

List all solutions of -4\sqrt{80-a^{2}}=-2a^{2}.

$$a=4 a=-4$$