Solve -1/2(frac{sqrt{17}-1}{2})^2-1/2(frac{sqrt{17}-1}{2})+3

$$- \frac { 1 } { 2 } ( \frac { \sqrt { 17 } - 1 } { 2 } ) ^ { 2 } - \frac { 1 } { 2 } ( \frac { \sqrt { 17 } - 1 } { 2 } ) + 3$$

Evaluate

$$1$$

Short Solution Steps

To raise \frac{\sqrt{17}-1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$-\frac{1}{2}\times \frac{\left(\sqrt{17}-1\right)^{2}}{2^{2}}-\frac{1}{2}\times \frac{\sqrt{17}-1}{2}+3$$

Multiply -\frac{1}{2} times \frac{\left(\sqrt{17}-1\right)^{2}}{2^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(\sqrt{17}-1\right)^{2}}{2\times 2^{2}}-\frac{1}{2}\times \frac{\sqrt{17}-1}{2}+3$$

Multiply \frac{1}{2} times \frac{\sqrt{17}-1}{2} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(\sqrt{17}-1\right)^{2}}{2\times 2^{2}}-\frac{\sqrt{17}-1}{2\times 2}+3$$

Multiply 2 and 2 to get 4.

$$\frac{-\left(\sqrt{17}-1\right)^{2}}{2\times 2^{2}}-\frac{\sqrt{17}-1}{4}+3$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\times 2^{2} and 4 is 8. Multiply \frac{\sqrt{17}-1}{4} times \frac{2}{2}.

$$\frac{-\left(\sqrt{17}-1\right)^{2}}{8}-\frac{2\left(\sqrt{17}-1\right)}{8}+3$$

Since \frac{-\left(\sqrt{17}-1\right)^{2}}{8} and \frac{2\left(\sqrt{17}-1\right)}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-\left(\sqrt{17}-1\right)^{2}-2\left(\sqrt{17}-1\right)}{8}+3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{8}{8}.

$$\frac{-\left(\sqrt{17}-1\right)^{2}-2\left(\sqrt{17}-1\right)}{8}+\frac{3\times 8}{8}$$

Since \frac{-\left(\sqrt{17}-1\right)^{2}-2\left(\sqrt{17}-1\right)}{8} and \frac{3\times 8}{8} have the same denominator, add them by adding their numerators.

$$\frac{-\left(\sqrt{17}-1\right)^{2}-2\left(\sqrt{17}-1\right)+3\times 8}{8}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{17}-1\right)^{2}.

$$\frac{-\left(\left(\sqrt{17}\right)^{2}-2\sqrt{17}+1\right)-2\left(\sqrt{17}-1\right)}{8}+3$$

The square of \sqrt{17} is 17.

$$\frac{-\left(17-2\sqrt{17}+1\right)-2\left(\sqrt{17}-1\right)}{8}+3$$

Add 17 and 1 to get 18.

$$\frac{-\left(18-2\sqrt{17}\right)-2\left(\sqrt{17}-1\right)}{8}+3$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{8}{8}.

$$\frac{-\left(18-2\sqrt{17}\right)-2\left(\sqrt{17}-1\right)}{8}+\frac{3\times 8}{8}$$

Since \frac{-\left(18-2\sqrt{17}\right)-2\left(\sqrt{17}-1\right)}{8} and \frac{3\times 8}{8} have the same denominator, add them by adding their numerators.

$$\frac{-\left(18-2\sqrt{17}\right)-2\left(\sqrt{17}-1\right)+3\times 8}{8}$$

Do the multiplications in -\left(18-2\sqrt{17}\right)-2\left(\sqrt{17}-1\right)+3\times 8.

$$\frac{-18+2\sqrt{17}-2\sqrt{17}+2+24}{8}$$

Do the calculations in -18+2\sqrt{17}-2\sqrt{17}+2+24.

$$\frac{8}{8}$$

Factor

$$1$$