Question

is negative square root of 28 to the power of 3 < or > than negative square root of 7

Answer 1

To compare the expressions \( (-\sqrt{28})^3 \) and \( -\sqrt{7} \), we can calculate each one step by step.

First, let's compute \( (-\sqrt{28})^3 \):

1. Calculate \( \sqrt{28} \):
\[
\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}
\]

2. Now compute \( (-\sqrt{28})^3 \):
\[
(-\sqrt{28})^3 = (-2\sqrt{7})^3 = -8 \cdot 7^{3/2} = -8 \cdot 7\sqrt{7}
\]

Next, compute \( -\sqrt{7} \).

Now, we need to compare \( -8 \cdot 7\sqrt{7} \) and \( -\sqrt{7} \). This comparison is easier if we divide both sides of the inequality by \( -\sqrt{7} \). Since both sides are negative, dividing by \( -\sqrt{7} \) flips the inequality.

Comparing:

1. \( -8 \cdot 7\sqrt{7} \div -\sqrt{7} \) simplifies to:
\[
8 \cdot 7 = 56
\]

2. And on the right-hand side:
\[
-\sqrt{7} \div -\sqrt{7} = 1
\]

So, we have:
\[
56 > 1
\]

This means:
\[
-\sqrt{28}^3 < -\sqrt{7}
\]

Therefore, we can conclude:
\[
(-\sqrt{28})^3 < -\sqrt{7}
\]