ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: TylerD9 on April 15, 2019, 04:29:43 pm
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Hello,
Could someone please explain how to rationalise the denominator for the following:
1/(root(2) + root(3) + root(5))
1/(1-2^(1/3)).
Thank you :)
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These look like super challenging questions. I wouldn't expect anything close to this to be on the spec exam!
For the first one, you will want to remove each of the square roots one by one in the denominator. The first step could be to multiply the numerator and denominator by (sqrt(2) + sqrt(3)) - sqrt(5). That would remove the sqrt(5) from the denominator. Next, you'll be working with sqrt(2) and sqrt(3) in the denominator. So you'll have a denominator in the form of a*sqrt(2) + b*sqrt(3) + c, where a,b,c are integers. At this stage, you'd multiply numerator and denominator by (a*sqrt(2) - b*sqrt(3) + c), which will remove the sqrt(3). If you're playing around with such hard questions, I'm sure you can figure out the last step yourself! The denominator should be only in terms of sqrt(2) by this stage.
For the second question, you have to apply a pretty clever little trick
Multiply the denominator and numerator by (2^(2/3)+2^(1/3)+1). Multiply it out and watch the magic happen!
This trick comes from the factorization of 1-a^3=(a-1)(a^2+a+1). Here, we set a=2^(1/3)
Hope this helps!
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Actually, using that method, the computations for the first one isn't too bad. This is more or less a consequence of the fact that \(2+3=5\).
\begin{align*} \frac{1}{\sqrt2+\sqrt3+\sqrt5}&= \frac{\sqrt2+\sqrt3-\sqrt5}{\left(\sqrt2+\sqrt3+\sqrt5 \right)\left(\sqrt2+\sqrt3-\sqrt5\right)} \\ &= \frac{\sqrt2+\sqrt3-\sqrt5}{\left(\sqrt2+\sqrt3\right)^2 - \left(\sqrt5\right)^2} \tag{diff. of two squares}\\ &= \frac{\sqrt2+\sqrt3-\sqrt5}{\left(2+2\sqrt6+3\right)-5}\tag{perfect square}\\ &= \frac{\sqrt2+\sqrt3-\sqrt5}{2\sqrt6}\\ &= \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}\end{align*}
(Of course, the hard part is still recognising to multiply with \( \sqrt2+\sqrt3-\sqrt5\) in the first place though.)
Remark: Interestingly, if we multiplied by \(\sqrt2-\sqrt3-\sqrt5 \), more effort would have been required.