Question

1. $-4sqrt{18}$
2. $-sqrt{63a^4}$
3. $sqrt{112xy^3}$
4. $sqrt{45m^2pq^4}$

Explanation:

Problem 6: Simplify \(-4\sqrt{18}\)

Step 1: Factor the radicand

Factor \(18\) into \(9\times2\), since \(9\) is a perfect square.
\(\sqrt{18}=\sqrt{9\times2}\)

Step 2: Use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\)

\(\sqrt{9\times2}=\sqrt{9}\cdot\sqrt{2}=3\sqrt{2}\)

Step 3: Multiply by the coefficient

Multiply \(-4\) with \(3\sqrt{2}\).
\(-4\times3\sqrt{2}=-12\sqrt{2}\)

Step 1: Factor the radicand

Factor \(63\) into \(9\times7\) and \(a^{4}\) is \((a^{2})^{2}\), a perfect square.
\(\sqrt{63a^{4}}=\sqrt{9\times7\times(a^{2})^{2}}\)

Step 2: Use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\)

\(\sqrt{9\times7\times(a^{2})^{2}}=\sqrt{9}\cdot\sqrt{7}\cdot\sqrt{(a^{2})^{2}} = 3a^{2}\sqrt{7}\)

Step 3: Apply the negative sign

Multiply by \(-1\).
\(-3a^{2}\sqrt{7}\)

Step 1: Factor the radicand

Factor \(112\) into \(16\times7\), \(y^{3}\) into \(y^{2}\times y\).
\(\sqrt{112xy^{3}}=\sqrt{16\times7\times x\times y^{2}\times y}\)

Step 2: Use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\)

\(\sqrt{16\times7\times x\times y^{2}\times y}=\sqrt{16}\cdot\sqrt{y^{2}}\cdot\sqrt{7xy}=4y\sqrt{7xy}\)

Answer:

\(-12\sqrt{2}\)

Problem 8: Simplify \(-\sqrt{63a^{4}}\)