Solve (a^2+1/2)^2-2(2^2+1/8)

$$( a ^ { 2 } + \frac { 1 } { 2 } ) ^ { 2 } - 2 ( 2 ^ { 2 } + \frac { 1 } { 8 } )$$

Evaluate

$$a^{4}+a^{2}-8$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+\frac{1}{2}\right)^{2}.

$$\left(a^{2}\right)^{2}+a^{2}+\frac{1}{4}-2\left(2^{2}+\frac{1}{8}\right)$$

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

$$a^{4}+a^{2}+\frac{1}{4}-2\left(2^{2}+\frac{1}{8}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{4}+a^{2}+\frac{1}{4}-2\left(4+\frac{1}{8}\right)$$

Add 4 and \frac{1}{8} to get \frac{33}{8}.

$$a^{4}+a^{2}+\frac{1}{4}-2\times \frac{33}{8}$$

Multiply 2 and \frac{33}{8} to get \frac{33}{4}.

$$a^{4}+a^{2}+\frac{1}{4}-\frac{33}{4}$$

Subtract \frac{33}{4} from \frac{1}{4} to get -8.

$$a^{4}+a^{2}-8$$

Expand

$$a^{4}+a^{2}-8$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a^{2}+\frac{1}{2}\right)^{2}.

$$\left(a^{2}\right)^{2}+a^{2}+\frac{1}{4}-2\left(2^{2}+\frac{1}{8}\right)$$

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

$$a^{4}+a^{2}+\frac{1}{4}-2\left(2^{2}+\frac{1}{8}\right)$$

Calculate 2 to the power of 2 and get 4.

$$a^{4}+a^{2}+\frac{1}{4}-2\left(4+\frac{1}{8}\right)$$

Add 4 and \frac{1}{8} to get \frac{33}{8}.

$$a^{4}+a^{2}+\frac{1}{4}-2\times \frac{33}{8}$$

Multiply 2 and \frac{33}{8} to get \frac{33}{4}.

$$a^{4}+a^{2}+\frac{1}{4}-\frac{33}{4}$$

Subtract \frac{33}{4} from \frac{1}{4} to get -8.

$$a^{4}+a^{2}-8$$