Question

express as a fraction in simplest form with a rational denominator: \\(\frac{-2}{-3 + \sqrt{15}}\\)

Explanation:

Step1: Identify the conjugate of the denominator

The denominator is \(-3 + \sqrt{15}\), so its conjugate is \(-3 - \sqrt{15}\). We multiply the numerator and the denominator by this conjugate to rationalize the denominator.

$$ \frac{-2}{-3 + \sqrt{15}} \times \frac{-3 - \sqrt{15}}{-3 - \sqrt{15}} $$

Step2: Multiply the numerators and the denominators

First, multiply the numerators: \((-2) \times (-3 - \sqrt{15}) = 6 + 2\sqrt{15}\)
Then, multiply the denominators using the difference of squares formula \((a + b)(a - b)=a^2 - b^2\), where \(a=-3\) and \(b = \sqrt{15}\):

$$ (-3)^2-(\sqrt{15})^2=9 - 15=-6 $$

So now we have the fraction \(\frac{6 + 2\sqrt{15}}{-6}\)

Step3: Simplify the fraction

We can factor out a 2 from the numerator: \(\frac{2(3+\sqrt{15})}{-6}\)
Then divide both the numerator and the denominator by 2: \(\frac{3+\sqrt{15}}{-3}\)
We can also write this as \(-\frac{3 + \sqrt{15}}{3}\) or simplify further by splitting the fraction: \(-1-\frac{\sqrt{15}}{3}\), but let's check the initial multiplication again for any errors. Wait, let's re - do the numerator multiplication: \((-2)\times(-3-\sqrt{15})=(-2)\times(-3)+(-2)\times(-\sqrt{15}) = 6 + 2\sqrt{15}\), denominator: \((-3)^2-(\sqrt{15})^2 = 9 - 15=-6\). So \(\frac{6 + 2\sqrt{15}}{-6}=\frac{2(3+\sqrt{15})}{-6}=\frac{3 + \sqrt{15}}{-3}=-\frac{3+\sqrt{15}}{3}\). Alternatively, we can factor out a negative sign from the denominator first. Let's try another approach:
Original fraction: \(\frac{-2}{-3+\sqrt{15}}=\frac{-2}{\sqrt{15}-3}\) (rearranging the denominator terms). Now the conjugate of \(\sqrt{15}-3\) is \(\sqrt{15}+3\). Multiply numerator and denominator by \(\sqrt{15}+3\):

$$ \frac{-2(\sqrt{15}+3)}{(\sqrt{15}-3)(\sqrt{15}+3)}=\frac{-2\sqrt{15}-6}{15 - 9}=\frac{-2\sqrt{15}-6}{6}=\frac{-2(\sqrt{15}+3)}{6}=\frac{-(\sqrt{15}+3)}{3}=-\frac{\sqrt{15}+3}{3} $$

Wait, there was a mistake in the first conjugate choice. The denominator is \(-3+\sqrt{15}=\sqrt{15}-3\), so the conjugate is \(\sqrt{15}+3\) (because for \(a + b\), conjugate is \(a - b\), here \(a=\sqrt{15}\), \(b = 3\)). So let's correct step 1:

Step1 (corrected): Identify the correct conjugate

The denominator is \(\sqrt{15}-3\) (since \(-3+\sqrt{15}=\sqrt{15}-3\)), so its conjugate is \(\sqrt{15}+3\). Multiply numerator and denominator by \(\sqrt{15}+3\):

$$ \frac{-2}{\sqrt{15}-3}\times\frac{\sqrt{15}+3}{\sqrt{15}+3} $$

Step2 (corrected): Multiply numerators and denominators

Numerator: \(-2\times(\sqrt{15}+3)=-2\sqrt{15}-6\)
Denominator: \((\sqrt{15})^2-3^2 = 15 - 9 = 6\)
So the fraction becomes \(\frac{-2\sqrt{15}-6}{6}\)

Step3 (corrected): Simplify the fraction

Factor out a - 2 from the numerator: \(\frac{-2(\sqrt{15}+3)}{6}\)
Divide numerator and denominator by 2: \(\frac{-(\sqrt{15}+3)}{3}=-\frac{\sqrt{15}+3}{3}\) or we can write it as \(\frac{- \sqrt{15}-3}{3}\) or split it as \(-1-\frac{\sqrt{15}}{3}\). Let's check the arithmetic again.
Wait, \((\sqrt{15}-3)(\sqrt{15}+3)=15 - 9 = 6\) (correct, difference of squares). Numerator: \(-2\times(\sqrt{15}+3)=-2\sqrt{15}-6\) (correct). Then \(\frac{-2\sqrt{15}-6}{6}=\frac{-2(\sqrt{15}+3)}{6}=\frac{-(\sqrt{15}+3)}{3}\) (dividing numerator and denominator by 2). Yes, that's correct.

Answer:

\(-\frac{3+\sqrt{15}}{3}\) (or \(\frac{-3 - \sqrt{15}}{3}\) or \(-1-\frac{\sqrt{15}}{3}\))