Solve 1^2+2*{left(1/4right)}^2-2*1-1/4-4

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

Evaluate

$$-\frac{41}{8}=-5.125$$

Solution Steps

Calculate 1 to the power of 2 and get 1.

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

Calculate \frac{1}{4} to the power of 2 and get \frac{1}{16}.

$$1+2\times \frac{1}{16}-2\times 1-\frac{1}{4}-4$$

Multiply 2 and \frac{1}{16} to get \frac{2}{16}.

$$1+\frac{2}{16}-2\times 1-\frac{1}{4}-4$$

Reduce the fraction \frac{2}{16} to lowest terms by extracting and canceling out 2.

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

Convert 1 to fraction \frac{8}{8}.

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

Since \frac{8}{8} and \frac{1}{8} have the same denominator, add them by adding their numerators.

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

Add 8 and 1 to get 9.

$$\frac{9}{8}-2\times 1-\frac{1}{4}-4$$

Multiply 2 and 1 to get 2.

$$\frac{9}{8}-2-\frac{1}{4}-4$$

Convert 2 to fraction \frac{16}{8}.

$$\frac{9}{8}-\frac{16}{8}-\frac{1}{4}-4$$

Since \frac{9}{8} and \frac{16}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{9-16}{8}-\frac{1}{4}-4$$

Subtract 16 from 9 to get -7.

$$-\frac{7}{8}-\frac{1}{4}-4$$

Least common multiple of 8 and 4 is 8. Convert -\frac{7}{8} and \frac{1}{4} to fractions with denominator 8.

$$-\frac{7}{8}-\frac{2}{8}-4$$

Since -\frac{7}{8} and \frac{2}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-7-2}{8}-4$$

Subtract 2 from -7 to get -9.

$$-\frac{9}{8}-4$$

Convert 4 to fraction \frac{32}{8}.

$$-\frac{9}{8}-\frac{32}{8}$$

Since -\frac{9}{8} and \frac{32}{8} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-9-32}{8}$$

Subtract 32 from -9 to get -41.

$$-\frac{41}{8}$$

Factor

$$-\frac{41}{8} = -5\frac{1}{8} = -5.125$$