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  
Senior Member Joined: Sep 2015 From: CA Posts: 1,207 Thanks: 615  Quote:
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  
Senior Member Joined: Jan 2015 From: London Posts: 102 Thanks: 2  Quote:
 
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  
May 9th, 2017, 11:41 AM  #8 
Senior Member Joined: Jan 2015 From: London Posts: 102 Thanks: 2  

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