Question

enter the correct answer in the box.
if ( x > 0 ), what is the product of ( 7sqrt{5x^3} cdot 9xsqrt{24x} ) in simplest radical form?

Explanation:

Step1: Multiply the coefficients and the variables outside the radicals

First, multiply the coefficients \(7\) and \(9\), and the variables outside the radicals (the \(x\) from \(9x\) and the \(x\) terms from the radicands will be handled later). So, \(7\times9 = 63\), and for the \(x\) terms, we have \(x\) (from \(9x\)) and we'll deal with the \(x\) in the radicand \(x^{3}\) and \(x\) in \(24x\) when simplifying the radicand. So far, we have \(63x\) (temporarily, before combining the radicands).

Step2: Multiply the radicands together

Now, multiply the radicands \(\sqrt{5x^{3}}\) and \(\sqrt{24x}\). Using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\), we get \(\sqrt{5x^{3}\cdot24x}\). Let's simplify the product inside the radical: \(5\times24 = 120\), and \(x^{3}\times x=x^{3 + 1}=x^{4}\). So the radicand becomes \(\sqrt{120x^{4}}\).

Step3: Simplify the radical \(\sqrt{120x^{4}}\)

We can factor \(120\) and \(x^{4}\) to simplify the square root. First, factor \(120\): \(120=4\times30\) (since \(4\) is a perfect square). And \(x^{4}=(x^{2})^{2}\), which is a perfect square. So, \(\sqrt{120x^{4}}=\sqrt{4\times30\times(x^{2})^{2}}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for non - negative \(a\) and \(b\)), we get \(\sqrt{4}\times\sqrt{(x^{2})^{2}}\times\sqrt{30}=2\times x^{2}\times\sqrt{30}=2x^{2}\sqrt{30}\).

Step4: Combine the results from Step1 and Step3

We had \(63x\) from Step1 (before simplifying the radical) and we found that the simplified radical is \(2x^{2}\sqrt{30}\)? Wait, no, wait. Wait, in Step1, we had \(7\times9 = 63\) and the \(x\) from \(9x\), and when we multiply the radicands, we have to consider the \(x\) in the first radicand. Let's re - do Step1 and Step2 more carefully.

The original expression is \(7\sqrt{5x^{3}}\cdot9x\sqrt{24x}\). First, multiply the coefficients: \(7\times9 = 63\). Then, multiply the \(x\) terms: \(x\) (from \(9x\)) and \(x^{3/2}\) (since \(\sqrt{x^{3}}=x^{3/2}\)) and \(x^{1/2}\) (since \(\sqrt{x}=x^{1/2}\))? Wait, maybe a better way:

\(7\sqrt{5x^{3}}\cdot9x\sqrt{24x}=(7\times9)\times(x)\times\sqrt{5x^{3}\times24x}\)

\(= 63x\times\sqrt{120x^{4}}\)

Now, \(\sqrt{120x^{4}}=\sqrt{4\times30\times x^{4}}=\sqrt{4}\times\sqrt{x^{4}}\times\sqrt{30}=2\times x^{2}\times\sqrt{30}\)

So now, multiply \(63x\) with \(2x^{2}\sqrt{30}\): \(63\times2 = 126\), and \(x\times x^{2}=x^{3}\). So we get \(126x^{3}\sqrt{30}\)? Wait, no, wait, I made a mistake in Step1. Wait, the first radical is \(\sqrt{5x^{3}}\), which can be written as \(\sqrt{5x^{2}\cdot x}=x\sqrt{5x}\) (since \(x\gt0\), \(\sqrt{x^{2}} = x\)). The second radical is \(9x\sqrt{24x}\). Let's rewrite the original expression by simplifying each radical first:

\(7\sqrt{5x^{3}}=7\sqrt{x^{2}\cdot5x}=7x\sqrt{5x}\) (because \(x\gt0\), so \(\sqrt{x^{2}}=x\))

\(9x\sqrt{24x}=9x\sqrt{4\times6x}=9x\times2\sqrt{6x}=18x\sqrt{6x}\)

Now, multiply the two simplified expressions: \((7x\sqrt{5x})\cdot(18x\sqrt{6x})\)

Multiply the coefficients: \(7\times18 = 126\)

Multiply the \(x\) terms: \(x\times x=x^{2}\)

Multiply the radicals: \(\sqrt{5x}\cdot\sqrt{6x}=\sqrt{5x\times6x}=\sqrt{30x^{2}}\)

Simplify \(\sqrt{30x^{2}}\): since \(x\gt0\), \(\sqrt{x^{2}} = x\), so \(\sqrt{30x^{2}}=x\sqrt{30}\)

Now, combine all the parts: \(126\times x^{2}\times x\sqrt{30}=126x^{3}\sqrt{30}\)? Wait, no, wait:

Wait, \((7x\sqrt{5x})\cdot(18x\sqrt{6x})=(7\times18)\times(x\times x)\times(\sqrt{5x}\times\sqrt{6x})\)

\(= 126x^{2}\times\sqrt{30x^{2}}\)

\(=126x^{2}\times x\sqrt{30}\) (because \(\sqrt{30x^{2}}=x\sqrt{30}\) as \(x\gt0\))

\(=126x^{3}\sqrt{30}\)

Wait, but let's check again. Wait, when we multiply \(\sqrt{5x}\) and \(\sqrt{6x}\), we have \(\sqrt{(5x)(6x)}=\sqrt{30x^{2}}=x\sqrt{30}\) (since \(x\gt0\)). Then, the \(x\) from \(7x\) and \(x\) from \(18x\) multiply to \(x^{2}\), and then we multiply by \(x\) from \(\sqrt{30x^{2}}\) to get \(x^{3}\). And \(7\times18 = 126\). So the product is \(126x^{3}\sqrt{30}\)? Wait, no, wait, I think I messed up the initial simplification. Let's go back to the first method:

Original expression: \(7\sqrt{5x^{3}}\cdot9x\sqrt{24x}\)

Multiply the coefficients: \(7\times9 = 63\)

Multiply the \(x\) terms: \(x\) (from \(9x\)) and \(x^{3/2}\) (from \(\sqrt{x^{3}}\)) and \(x^{1/2}\) (from \(\sqrt{x}\))? Wait, \(\sqrt{5x^{3}}=5^{1/2}x^{3/2}\) and \(\sqrt{24x}=24^{1/2}x^{1/2}\), and we have a factor of \(x\) from \(9x\). So the exponent of \(x\) is \(3/2+1/2 + 1=3\) (since \(x = x^{1}\)). Then the product of the radicands' coefficients: \(5\times24 = 120\). So we have \(63x^{3}\sqrt{120}\). Now, simplify \(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\). So \(63x^{3}\times2\sqrt{30}=126x^{3}\sqrt{30}\). Yes, that's correct.

Answer:

\(126x^{3}\sqrt{30}\)