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PS algebra

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young_gun
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PS algebra [#permalink] New post   18 Dec 2007, 13:31
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what's the easiest way to solve these types:

5(t^2)-14t-24=0



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Re: PS algebra [#permalink] New post   18 Dec 2007, 13:52
i think there is always more than one way, but i did this:

5t^2-14t-24=0
5t^2-14t=24
t(5t-14)=24
t=24 and 5t=14, t=14/5



young_gun wrote:
what's the easiest way to solve these types:

5(t^2)-14t-24=0
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Re: PS algebra [#permalink] New post   18 Dec 2007, 13:56
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michaelny2001 wrote:
t(5t-14)=24
t=24 and 5t=14, t=14/5


:shock:
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Re: PS algebra [#permalink] New post   18 Dec 2007, 20:36
walker wrote:
michaelny2001 wrote:
t(5t-14)=24
t=24 and 5t=14, t=14/5


:shock:


t(5t-14)=24
t=24 and 5t=14, t=14/5
this is wrong i guess

Way to solve simple quadratic equations

x^2 - 5x + 6 = 0=> break the middle term such that its product is equal to the product of end terms

x^2 -3x -2x +6 = 0
x(x-3) - 2(x-3) = 0
(x-2)(x-3) = 0
Roots are 2 and 3 .

If equations are not simple apply direct formula for roots for equations like ax^2 + bx +c = 0
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Re: PS algebra [#permalink] New post   18 Dec 2007, 21:23
If I can't transform equation into the basic multiples, I use formula above.

But GMAT usually uses equations that are easily transformed into the form (a + b) * (a + c) or analogues
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Re: PS algebra [#permalink] New post   18 Dec 2007, 23:38
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There a general shortcut for equation like:

x^n+a*x^(n-1)+....+g=0

if g integer your roots probably will be among +/-factors of g.

For example:

x^2 -3x -2x +6 = 0

g=6 ==> {-6,-3,-2,-1,1,2,3,6} - both roots are in the set
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Re: PS algebra [#permalink] New post   18 Dec 2007, 23:46
young_gun wrote:
what's the easiest way to solve these types:

5(t^2)-14t-24=0


24*5 = 120; If we take -20 and 6 as factors then sum = -20+6 = -14

5t^2 - 20t + 6t - 24 = 0
5t(t-4)+6(t-4) = 0
(5t+6)(t-4) = 0
t=-6/5 or t=4
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Re: PS algebra [#permalink] New post   19 Dec 2007, 15:02
ashkrs wrote:
walker wrote:
michaelny2001 wrote:
t(5t-14)=24
t=24 and 5t=14, t=14/5


:shock:


t(5t-14)=24
t=24 and 5t=14, t=14/5
this is wrong i guess

Way to solve simple quadratic equations

x^2 - 5x + 6 = 0=> break the middle term such that its product is equal to the product of end terms

x^2 -3x -2x +6 = 0
x(x-3) - 2(x-3) = 0
(x-2)(x-3) = 0
Roots are 2 and 3 .

If equations are not simple apply direct formula for roots for equations like ax^2 + bx +c = 0


O damn highschool math. forgot all bout that quadratic formula thing.

What is it even called. Thats pretty handy actually.

essentially x=(14+/-26)/10 --> x=6/5 or 4
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Re: PS algebra [#permalink] New post   22 Dec 2007, 02:22
GK_Gmat wrote:
young_gun wrote:
what's the easiest way to solve these types:

5(t^2)-14t-24=0


24*5 = 120; If we take -20 and 6 as factors then sum = -20+6 = -14

5t^2 - 20t + 6t - 24 = 0
5t(t-4)+6(t-4) = 0
(5t+6)(t-4) = 0
t=-6/5 or t=4


This approach seems to be correct? Anyone who disagrees with this?



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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gmatclubot
Re: PS algebra [#permalink]
22 Dec 2007, 02:22
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