Solve (2sqrt{2}-7/8)^2-(2sqrt{2}+7/8)^2

$$( 2 \sqrt { 2 } - \frac { 7 } { 8 } ) ^ { 2 } - ( 2 \sqrt { 2 } + \frac { 7 } { 8 } ) ^ { 2 }$$

Evaluate

$$-7\sqrt{2}\approx -9.899494937$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-\frac{7}{8}\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4\times 2-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Add 8 and \frac{49}{64} to get \frac{561}{64}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{2}+\frac{7}{8}\right)^{2}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(4\left(\sqrt{2}\right)^{2}+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

The square of \sqrt{2} is 2.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(4\times 2+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

Multiply 4 and 2 to get 8.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(8+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

Add 8 and \frac{49}{64} to get \frac{561}{64}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(\frac{561}{64}+\frac{7}{2}\sqrt{2}\right)$$

To find the opposite of \frac{561}{64}+\frac{7}{2}\sqrt{2}, find the opposite of each term.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\frac{561}{64}-\frac{7}{2}\sqrt{2}$$

Subtract \frac{561}{64} from \frac{561}{64} to get 0.

$$-\frac{7}{2}\sqrt{2}-\frac{7}{2}\sqrt{2}$$

Combine -\frac{7}{2}\sqrt{2} and -\frac{7}{2}\sqrt{2} to get -7\sqrt{2}.

$$-7\sqrt{2}$$

Expand

$$-7\sqrt{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-\frac{7}{8}\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4\times 2-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$8-\frac{7}{2}\sqrt{2}+\frac{49}{64}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Add 8 and \frac{49}{64} to get \frac{561}{64}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(2\sqrt{2}+\frac{7}{8}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2\sqrt{2}+\frac{7}{8}\right)^{2}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(4\left(\sqrt{2}\right)^{2}+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

The square of \sqrt{2} is 2.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(4\times 2+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

Multiply 4 and 2 to get 8.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(8+\frac{7}{2}\sqrt{2}+\frac{49}{64}\right)$$

Add 8 and \frac{49}{64} to get \frac{561}{64}.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\left(\frac{561}{64}+\frac{7}{2}\sqrt{2}\right)$$

To find the opposite of \frac{561}{64}+\frac{7}{2}\sqrt{2}, find the opposite of each term.

$$\frac{561}{64}-\frac{7}{2}\sqrt{2}-\frac{561}{64}-\frac{7}{2}\sqrt{2}$$

Subtract \frac{561}{64} from \frac{561}{64} to get 0.

$$-\frac{7}{2}\sqrt{2}-\frac{7}{2}\sqrt{2}$$

Combine -\frac{7}{2}\sqrt{2} and -\frac{7}{2}\sqrt{2} to get -7\sqrt{2}.

$$-7\sqrt{2}$$