Question Video: Evaluating Numerical Expressions Involving Square and Cube Roots Mathematics

Evaluate ∛−27 + ∛−0.125 + √5(1/16).

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Video Transcript

Evaluate the cube root of negative 27 plus the cube root of negative 0.125 plus the square root of five and one sixteenth.

In this question, we are asked to evaluate an expression involving three terms. The first term is the cube root of an integer. The second term is the cube root of a decimal. And the third term is the square root of a mixed number.

To evaluate this expression, we can start by recalling that it is often easiest to evaluate the roots of noninteger values when they are written as fractions. So we will start by converting negative 0.125 and five and one sixteenth into fractions. If we do this, then we can calculate that negative 0.125 is negative one-eighth and five and one sixteenth is 81 over 16. This allows us to rewrite the expression as the cube root of negative 27 plus the cube root of negative one-eighth plus the square root of 81 over 16.

We now need to evaluate this expression, and we will do this by evaluating each term separately. Let’s start with the first term, the cube root of negative 27. We can evaluate this term by noting that negative 27 is a perfect cube. In particular, negative 27 is equal to negative three cubed. We can then evaluate this term by recalling that for any real number 𝑎, we have that the cube root of 𝑎 cubed is equal to 𝑎. So, the cube root of negative three cubed is equal to negative three.

We can evaluate the second term by using the laws of exponents. We can recall that for any real numbers 𝑎 and 𝑏, with 𝑏 not equal to zero, we have that the cube root of 𝑎 over 𝑏 is equal to the cube root of 𝑎 over the cube root of 𝑏. Since negative one-eighth is equal to negative one over eight, we can use this result with 𝑎 equal to negative one and 𝑏 equal to eight to rewrite the second term as the cube root of negative one over the cube root of eight. We can then evaluate this term by noting that negative one is equal to negative one cubed and eight is equal to two cubed. Therefore, the numerator is equal to negative one and the denominator is equal to two. So this term evaluates to give negative one-half.

We can apply a similar process to the third term. We know that the square root of 𝑎 over 𝑏 is equal to the square root of 𝑎 over the square root of 𝑏, though we do need to be careful since we need 𝑎 to be nonnegative and 𝑏 to be positive. We can apply this result with 𝑎 equal to 81 and 𝑏 equal to 16 to get root 81 over root 16. Finally, we can evaluate the square root of the numerator and denominator. We know that root 81 is nine and root 16 is four. So we have negative three minus one-half plus nine-quarters.

All that is left to do is evaluate the expression. We rewrite the terms to have the same denominator and evaluate to obtain negative five-quarters.

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