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May 9th, 2017, 10:12 AM   #1
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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.
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May 9th, 2017, 11:05 AM   #2
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Quote:
Originally Posted by jamesbrown View Post
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*}$
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May 9th, 2017, 11:06 AM   #3
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w = width
w(1 + sqrt(5)) = sqrt(80)

Did you try using above?
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May 9th, 2017, 11:19 AM   #4
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Quote:
Originally Posted by romsek View Post
$\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?
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May 9th, 2017, 11:26 AM   #5
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Romsek: $1,000,000.23 fine for giving full solution !
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May 9th, 2017, 11:33 AM   #6
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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:
$\displaystyle 3+2\sqrt{5}$
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May 9th, 2017, 11:35 AM   #7
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Quote:
Originally Posted by jamesbrown View Post
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.
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May 9th, 2017, 11:41 AM   #8
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Quote:
Originally Posted by romsek View Post
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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