Solve sqrt{19/16}+sqrt[3]{-8}-(1/2)^2

$$\sqrt { 1 \frac { 9 } { 16 } } + \sqrt[ 3 ] { - 8 } - ( \frac { 1 } { 2 } ) ^ { 2 }$$

Evaluate

$$-1$$

Solution Steps

Multiply 1 and 16 to get 16.

$$\sqrt{\frac{16+9}{16}}+\sqrt[3]{-8}-\left(\frac{1}{2}\right)^{2}$$

Add 16 and 9 to get 25.

$$\sqrt{\frac{25}{16}}+\sqrt[3]{-8}-\left(\frac{1}{2}\right)^{2}$$

Rewrite the square root of the division \frac{25}{16} as the division of square roots \frac{\sqrt{25}}{\sqrt{16}}. Take the square root of both numerator and denominator.

$$\frac{5}{4}+\sqrt[3]{-8}-\left(\frac{1}{2}\right)^{2}$$

Calculate \sqrt[3]{-8} and get -2.

$$\frac{5}{4}-2-\left(\frac{1}{2}\right)^{2}$$

Subtract 2 from \frac{5}{4} to get -\frac{3}{4}.

$$-\frac{3}{4}-\left(\frac{1}{2}\right)^{2}$$

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

$$-\frac{3}{4}-\frac{1}{4}$$

Subtract \frac{1}{4} from -\frac{3}{4} to get -1.

$$-1$$

Factor

$$-1$$