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PS algebra

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PS algebra [#permalink] New post 18 Dec 2007, 12:31
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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, 12:52
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, 12:56
Expert's post
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, 19:36
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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 [#permalink] New post 18 Dec 2007, 20:23
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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 [#permalink] New post 18 Dec 2007, 22:38
Expert's post
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, 22:46
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, 14:02
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, 01:22
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?
Re: PS algebra   [#permalink] 22 Dec 2007, 01:22
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