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May 4th, 2016, 10:54 AM   #1
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Completing the square? I think?

Hi, I'm really stuck on this question could someone please help me out?
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May 4th, 2016, 11:28 AM   #2
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Use either "completing the square" or the "quadratic formula" to solve the equation. Either is fairly simple.
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May 4th, 2016, 11:42 AM   #3
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Quote:
Originally Posted by Country Boy View Post
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?
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May 4th, 2016, 11:47 AM   #4
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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?
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May 4th, 2016, 12:29 PM   #5
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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?
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May 4th, 2016, 01:06 PM   #6
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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.
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May 4th, 2016, 02:10 PM   #7
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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.
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May 5th, 2016, 08:13 AM   #8
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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.
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May 5th, 2016, 08:30 AM   #9
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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.
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May 5th, 2016, 09:00 AM   #10
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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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