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 May 4th, 2016, 10:54 AM #1 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 Completing the square? I think? Hi, I'm really stuck on this question could someone please help me out? Imgur: The most awesome images on the Internet
 May 4th, 2016, 11:28 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,159 Thanks: 866 Use either "completing the square" or the "quadratic formula" to solve the equation. Either is fairly simple.
May 4th, 2016, 11:42 AM   #3
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Quote:
 Originally Posted by Country Boy Use either "completing the square" or the "quadratic formula" to solve the equation. Either is fairly simple.
What do I do for question a though?

 May 4th, 2016, 11:47 AM #4 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 So I have to work out that x=20 and x=3.3 for a and this is the answer for it, just to work it out?
May 4th, 2016, 12:29 PM   #5
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Quote:
 Originally Posted by jamesbrown What do I do for question a though?
adding 10 meters to the width = x+10

adding 20 meters to the length = 3x+20

area of the extended playground is double the original ...

original area = $3x^2$, doubled = $6x^2$

$(x+10)(3x+20) = 6x^2$

expand the left side, combine like terms & set = 0 ...

$3x^2 + 50x + 200 = 6x^2$

$0 = 3x^2 - 50x - 200$

kapish?

 May 4th, 2016, 01:06 PM #6 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 But for b it says calculate the area of the original playground; which x value do I use? Last edited by skipjack; May 5th, 2016 at 11:56 AM.
May 4th, 2016, 02:10 PM   #7
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Quote:
 Originally Posted by jamesbrown But for b it says calculate the area of the original playground; which x value do I use?
Which $x$? There is only one $x$ that represents the original width.

Solve for $x$ from the doubling equation $3x^2-50x-200 = 0$ ...

(fyi, $3x^2-50x-200$ will factor)

... now determine the original area by evaluating the value of $3x^2$ using what you got for the one and only $x$.

Last edited by skipjack; May 5th, 2016 at 11:57 AM.

 May 5th, 2016, 08:13 AM #8 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 How am I supposed to work out x? I thought you had to do the completing the square method. Last edited by skipjack; May 5th, 2016 at 11:58 AM.
May 5th, 2016, 08:30 AM   #9
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Quote:
 Originally Posted by jamesbrown How am I supposed to work out x? I thought you had to do the completing the square method.
What's stopping you? Complete the square if that is your method of choice. I'll even get you started ...

$3x^2 - 50x - 200 = 0$

$3x^2 - 50x = 200$

$x^2 - \dfrac{50}{3}x = \dfrac{200}{3}$

$x^2 - \dfrac{50}{3}x + \left(\dfrac{25}{3}\right)^2 = \dfrac{200}{3}+ \left(\dfrac{25}{3}\right)^2$

finish it ...

Last edited by skipjack; May 5th, 2016 at 11:58 AM.

 May 5th, 2016, 09:00 AM #10 Senior Member   Joined: Jan 2015 From: London Posts: 108 Thanks: 2 I get x=20 or x=-10/3, so would I use 20 and then do 3*20=60 60*20=1200 so the area is 1200m^2? Is this right? Last edited by skipjack; May 5th, 2016 at 11:59 AM.

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