Question

evaluate the following square root expression. assume x > 0. \\(\sqrt{-18x^9}\\)

Explanation:

Step1: Analyze the radicand

The radicand is \(-18x^9\). Since \(x>0\), \(x^9>0\), so \(-18x^9 = -1\times18\times x^9\). We know that for real - number square roots, the radicand must be non - negative. But here the radicand \(-18x^9<0\) (because \(x > 0\) implies \(x^9>0\) and \(-18<0\), so their product is negative). However, we can use the concept of imaginary numbers. Recall that \(\sqrt{-a}=\sqrt{a}\times i\) where \(a>0\) and \(i=\sqrt{-1}\) is the imaginary unit. Also, we can simplify the square root of the positive part using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a\geq0,b\geq0\)) and \(\sqrt{x^n}=x^{\frac{n}{2}}\) for \(x\geq0\).

First, rewrite \(-18x^9\) as \(-1\times9\times2\times x^8\times x\).

Step2: Simplify the square root

We know that \(\sqrt{-18x^9}=\sqrt{-1\times18\times x^9}\). Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a = - 1\) and \(b = 18x^9\), and extending the property to complex numbers), we have \(\sqrt{-1\times18\times x^9}=\sqrt{-1}\cdot\sqrt{18x^9}\).

Since \(\sqrt{-1}=i\), we now simplify \(\sqrt{18x^9}\). We can factor \(18 = 9\times2\) and \(x^9=x^8\times x\). Then \(\sqrt{18x^9}=\sqrt{9\times2\times x^8\times x}\).

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) again, we get \(\sqrt{9}\cdot\sqrt{2}\cdot\sqrt{x^8}\cdot\sqrt{x}\).

We know that \(\sqrt{9} = 3\), \(\sqrt{x^8}=x^{\frac{8}{2}}=x^4\) (because \(x>0\)), so \(\sqrt{9}\cdot\sqrt{2}\cdot\sqrt{x^8}\cdot\sqrt{x}=3x^4\sqrt{2x}\).

Putting it all together, \(\sqrt{-18x^9}=i\times3x^4\sqrt{2x}=3x^4i\sqrt{2x}\) or \(3\sqrt{2}x^4\sqrt{x}i = 3x^{4 + \frac{1}{2}}i\sqrt{2}=3x^{\frac{9}{2}}i\sqrt{2}\) (but the more standard form is \(3x^4\sqrt{2x}i\)).

Wait, let's check the exponent of \(x\) again. \(x^9=x^{8 + 1}\), so \(\sqrt{x^9}=\sqrt{x^8\cdot x}=\sqrt{x^8}\cdot\sqrt{x}=x^4\sqrt{x}\). And \(\sqrt{18}=\sqrt{9\times2}=3\sqrt{2}\). So \(\sqrt{18x^9}=3\sqrt{2}x^4\sqrt{x}\). Then \(\sqrt{-18x^9}=i\times3\sqrt{2}x^4\sqrt{x}=3x^4\sqrt{2x}i\) (since \(\sqrt{2}\cdot\sqrt{x}=\sqrt{2x}\)).

Answer:

\(3x^4\sqrt{2x}i\) (or in a more factored form, the simplified form of the square root of the negative expression in the complex number system is \(3x^4i\sqrt{2x}\))