Question

simplify.
$6x^2\sqrt{72x} - \sqrt{18x^5}$
assume that the variable represents a positive real number.

Explanation:

Step1: Simplify the first radical

Simplify \( \sqrt{72x} \). We can factor 72 as \( 36\times2 \), so \( \sqrt{72x}=\sqrt{36\times2x} \). Using the property \( \sqrt{ab}=\sqrt{a}\sqrt{b} \) (where \( a = 36 \) and \( b = 2x \)), we get \( \sqrt{36}\sqrt{2x}=6\sqrt{2x} \). Then the first term \( 6x^{2}\sqrt{72x} \) becomes \( 6x^{2}\times6\sqrt{2x}=36x^{2}\sqrt{2x} \).

Step2: Simplify the second radical

Simplify \( \sqrt{18x^{5}} \). Factor 18 as \( 9\times2 \) and \( x^{5} \) as \( x^{4}\times x \). So \( \sqrt{18x^{5}}=\sqrt{9\times2\times x^{4}\times x} \). Using the property \( \sqrt{ab}=\sqrt{a}\sqrt{b} \) again, we have \( \sqrt{9}\sqrt{x^{4}}\sqrt{2x}=3x^{2}\sqrt{2x} \) (since \( \sqrt{x^{4}} = x^{2} \) for positive \( x \)).

Step3: Subtract the two simplified radicals

Now we have \( 36x^{2}\sqrt{2x}-3x^{2}\sqrt{2x} \). Since the radicals are the same, we can combine like terms. Subtract the coefficients: \( (36 - 3)x^{2}\sqrt{2x}=33x^{2}\sqrt{2x} \).

Answer:

\( 33x^{2}\sqrt{2x} \)