Question

multiply. assume q is greater than or equal to zero, and write your answer in simplest form. \\(sqrt{77q^{5}} cdot sqrt{98q^{4}}\\)

Explanation:

Step1: Multiply the radicands

Use the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\). So, \(\sqrt{77q^{5}}\cdot\sqrt{98q^{4}}=\sqrt{(77q^{5})(98q^{4})}\).
Calculate the product of the coefficients and the variables: \(77\times98 = 7546\), and \(q^{5}\times q^{4}=q^{5 + 4}=q^{9}\). So we have \(\sqrt{7546q^{9}}\).

Step2: Factor the radicand

Factor \(7546\) and \(q^{9}\) into perfect - square factors and non - perfect - square factors.
First, factor \(7546\): \(7546=49\times154 = 49\times14\times11=7^{2}\times2\times7\times11 = 7^{3}\times2\times11\)? Wait, no, let's do it correctly. \(77 = 7\times11\) and \(98 = 7\times14=7\times2\times7\). So \(77\times98=(7\times11)\times(7\times2\times7)=7^{3}\times2\times11\). And \(q^{9}=q^{8}\times q=(q^{4})^{2}\times q\).
So \(\sqrt{7546q^{9}}=\sqrt{7^{3}\times2\times11\times q^{8}\times q}\).
We can rewrite \(7^{3}\) as \(7^{2}\times7\) and \(q^{8}=(q^{4})^{2}\). Then \(\sqrt{7^{2}\times7\times2\times11\times(q^{4})^{2}\times q}\).

Step3: Simplify the square root

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)), we get:
\(\sqrt{7^{2}}\times\sqrt{(q^{4})^{2}}\times\sqrt{7\times2\times11\times q}\)
Since \(\sqrt{7^{2}} = 7\) and \(\sqrt{(q^{4})^{2}}=q^{4}\) (because \(q\geq0\)), we have:
\(7\times q^{4}\times\sqrt{154q}\) (because \(7\times2\times11 = 154\))
\(7q^{4}\sqrt{154q}\)? Wait, no, let's go back. Wait, \(77\times98=(7\times11)\times(7\times14)=(7\times11)\times(7\times2\times7)=7^{3}\times2\times11\). And \(q^{5}\times q^{4}=q^{9}\).
Wait, another way:
\(\sqrt{77q^{5}}\cdot\sqrt{98q^{4}}=\sqrt{77\times98\times q^{5 + 4}}=\sqrt{7546q^{9}}\)
Factor \(7546\): \(7546 = 49\times154=7^{2}\times154\), and \(q^{9}=q^{8}\times q=(q^{4})^{2}\times q\)
So \(\sqrt{7^{2}\times154\times(q^{4})^{2}\times q}=\sqrt{7^{2}}\times\sqrt{(q^{4})^{2}}\times\sqrt{154q}=7\times q^{4}\times\sqrt{154q}=7q^{4}\sqrt{154q}\)? Wait, no, \(77\times98 = 7\times11\times7\times14=7\times11\times7\times2\times7 = 7^{3}\times2\times11\), \(q^{5}\times q^{4}=q^{9}=q^{8}\times q=(q^{4})^{2}\times q\)
So \(\sqrt{7^{3}\times2\times11\times(q^{4})^{2}\times q}=\sqrt{7^{2}\times7\times2\times11\times(q^{4})^{2}\times q}=7\times q^{4}\times\sqrt{7\times2\times11\times q}=7q^{4}\sqrt{154q}\)? Wait, \(7\times2\times11 = 154\), yes. But wait, let's check the multiplication of the coefficients again. \(77\times98=(70 + 7)\times(90+8)=70\times90+70\times8 + 7\times90+7\times8=6300+560 + 630+56=6300+560=6860;6860 + 630=7490;7490+56 = 7546\). And \(7546\div49 = 154\) (since \(49\times154 = 49\times(150 + 4)=49\times150+49\times4=7350 + 196 = 7546\)).
Wait, but let's re - express the original problem:
\(\sqrt{77q^{5}}\cdot\sqrt{98q^{4}}=\sqrt{77}\cdot\sqrt{q^{5}}\cdot\sqrt{98}\cdot\sqrt{q^{4}}\)
\(=\sqrt{77\times98}\cdot\sqrt{q^{5}\times q^{4}}\)
\(=\sqrt{7\times11\times7\times14}\cdot\sqrt{q^{9}}\)
\(=\sqrt{7^{2}\times11\times14}\cdot\sqrt{q^{8}\times q}\)
\(=7\sqrt{154}\cdot q^{4}\sqrt{q}\)
\(=7q^{4}\sqrt{154q}\)
Wait, no, \(11\times14 = 154\), and \(q^{5}\times q^{4}=q^{9}=q^{8}\times q=(q^{4})^{2}\times q\). So \(\sqrt{q^{9}}=q^{4}\sqrt{q}\), and \(\sqrt{77\times98}=\sqrt{7\times11\times7\times14}=\sqrt{7^{2}\times11\times14}=7\sqrt{154}\). So multiplying them together gives \(7q^{4}\sqrt{154q}\). Wait, but let's check the coefficient again. \(77\times98=(7\times11)\times(7\times14)=7^{2}\times11\times14\), so \(\sqrt{7^{2}\times11\times14}=7\sqrt{154}\), and \(\sqrt{q^{5}\times q^{4}}=\sqrt{q^{9}}=q^{4}\sqrt{q}\). So the product is \(7q^{4}\sqrt{154q}\). Wait, but maybe we made a mistake in factoring. Let's try another approach.
\(77q^{5}=7\times11\times q^{4}\times q\) and \(98q^{4}=49\times2\times q^{4}=7^{2}\times2\times q^{4}\)
Then \(\sqrt{77q^{5}}\cdot\sqrt{98q^{4}}=\sqrt{7\times11\times q^{4}\times q}\cdot\sqrt{7^{2}\times2\times q^{4}}\)
Using \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we have:
\(\sqrt{(7\times11\times q^{4}\times q)\times(7^{2}\times2\times q^{4})}=\sqrt{7^{3}\times2\times11\times q^{8}\times q}\)
\(=\sqrt{7^{2}\times7\times2\times11\times(q^{4})^{2}\times q}\)
\(=7\times q^{4}\times\sqrt{7\times2\times11\times q}\)
\(=7q^{4}\sqrt{154q}\)

Answer:

\(7q^{4}\sqrt{154q}\)