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 May 9th, 2017, 10:12 AM #1 Senior Member   Joined: Jan 2015 From: London Posts: 102 Thanks: 2 Surd's Help! Write $\displaystyle \sqrt{80}$ in the form$\displaystyle c \sqrt{5}$, where c is a positive constant. So the answer is $\displaystyle 4\sqrt{5}$ But I dont know how to do surd questions like this, where you have to leave in a certain form. Question: A rectangle R has a length of $\displaystyle (1 + \sqrt{5})cm$ and an area of $\displaystyle \sqrt{80}cm^2$. Calculate the width of R in cm. Express your answer in the form $\displaystyle p + q\sqrt{5}$, where $\displaystyle p$ and $\displaystyle q$ are integers to be found. Last edited by jamesbrown; May 9th, 2017 at 10:18 AM.
May 9th, 2017, 11:05 AM   #2
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 Originally Posted by jamesbrown Write $\displaystyle \sqrt{80}$ in the form$\displaystyle c \sqrt{5}$, where c is a positive constant. So the answer is $\displaystyle 4\sqrt{5}$ But I dont know how to do surd questions like this, where you have to leave in a certain form. Question: A rectangle R has a length of $\displaystyle (1 + \sqrt{5})cm$ and an area of $\displaystyle \sqrt{80}cm^2$. Calculate the width of R in cm. Express your answer in the form $\displaystyle p + q\sqrt{5}$, where $\displaystyle p$ and $\displaystyle q$ are integers to be found.
\begin {align*} w &= \dfrac A \ell \\ \\ &= \dfrac{\sqrt{80}}{1+\sqrt{5}} \\ \\ &= \dfrac{\sqrt{80}}{1+\sqrt{5}} \dfrac{1-\sqrt{5}}{1-\sqrt{5}} \\ \\ &=\dfrac{\sqrt{80}(1-\sqrt{5})}{-4} \\ \\ &=-\dfrac{4\sqrt{5}(1-\sqrt{5})}{4} \\ \\ &=-(\sqrt{5}-5) \\ \\ &=5-\sqrt{5} \end{align*}

 May 9th, 2017, 11:06 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,083 Thanks: 624 w = width w(1 + sqrt(5)) = sqrt(80) Did you try using above?
May 9th, 2017, 11:19 AM   #4
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 Originally Posted by romsek \begin {align*} w &= \dfrac A \ell \\ \\ &= \dfrac{\sqrt{80}}{1+\sqrt{5}} \\ \\ &= \dfrac{\sqrt{80}}{1+\sqrt{5}} \dfrac{1-\sqrt{5}}{1-\sqrt{5}} \\ \\ &=\dfrac{\sqrt{80}(1-\sqrt{5})}{-4} \\ \\ &=-\dfrac{4\sqrt{5}(1-\sqrt{5})}{4} \\ \\ &=-(\sqrt{5}-5) \\ \\ &=5-\sqrt{5} \end{align*}
Thanks do you always have to rationalise when it says leave in form $\displaystyle x+y\sqrt{c}$ or whateva?

 May 9th, 2017, 11:26 AM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 9,083 Thanks: 624 Romsek: $1,000,000.23 fine for giving full solution !  May 9th, 2017, 11:33 AM #6 Senior Member Joined: Jan 2015 From: London Posts: 102 Thanks: 2 Ok so if I were to simplify this:$\displaystyle 7+\sqrt{5}\frac\sqrt{5}-1$in the form$\displaystyle a+b\sqrt{5}$I would rationalise first, so:$\displaystyle 7+\sqrt{5}\frac\sqrt{5}-1 * \sqrt{5}+1\frac\sqrt{5}+1\displaystyle 7\sqrt{5}+\sqrt[5}\sqrt{5}+\sqrt{5}+7\frac\sqrt{5}\sqrt{5}-\sqrt{5}+\sqrt{5}-1\displaystyle 12+8\sqrt{5}\frac4$Goes to:$\displaystyle 3+2\sqrt{5}$May 9th, 2017, 11:35 AM #7 Senior Member Joined: Sep 2015 From: CA Posts: 1,207 Thanks: 615 Quote:  Originally Posted by jamesbrown Thanks do you always have to rationalise when it says leave in form$\displaystyle x+y\sqrt{c}\$ or whateva?
the idea is to get radicals out of the denominator.

No radicals in the denominator is a sort of standard form.

May 9th, 2017, 11:41 AM   #8
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 Originally Posted by romsek the idea is to get radicals out of the denominator. No radicals in the denominator is a sort of standard form.
Thanks for this!

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