Question

Can someone help me simplify the radical equation please? I got stuck on a step and I'm not sure what I'm doing wrong.
√40x^4 / 3^√-x^3

Answer 1

Answer:

$-2x \sqrt{10}$

Step-by-step explanation:

To simplify the expression $\dfrac{\sqrt{40x^4}}{\sqrt[3]{-x^3}}$, we can apply the rules of indices.

Square Root Rule:

• $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

Cube Root Rule:

• $\sqrt[3]{a \cdot b} = \sqrt[3]{a} \cdot \sqrt[3]{b}$

Given expression:

$\dfrac{\sqrt{40x^4}}{\sqrt[3]{-x^3}}$

Apply the rules of indices:

$\dfrac{\sqrt{4 \cdot 10 \cdot x^2 \cdot x^2}}{\sqrt[3]{(-x)^3}}$

$\dfrac{\sqrt{(2 \cdot 2) \cdot 10 \cdot x^2 \cdot x^2}}{\sqrt[3]{(-1) \cdot x \cdot x \cdot x}}$

$\dfrac{(2 \cdot x \cdot x) \cdot \sqrt{10}}{(-1) \cdot x}$

$\dfrac{2x^2 \cdot \sqrt{10}}{-x}$

$\dfrac{2x^2}{-x} \cdot \sqrt{10}$

$-2x^{2-1} \cdot \sqrt{10}$

$-2x \cdot \sqrt{10}$

$-2x \sqrt{10}$

Therefore, the simplified expression is $-2x \sqrt{10}$.