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what's the easiest way to solve these types:
5(t^2)-14t-24=0

Manager

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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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michaelny2001 wrote:

t(5t-14)=24 t=24 and 5t=14, t=14/5

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walker wrote:

michaelny2001 wrote:

t(5t-14)=24 t=24 and 5t=14, t=14/5

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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Manager

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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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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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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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ashkrs wrote:

walker wrote:

michaelny2001 wrote:

t(5t-14)=24 t=24 and 5t=14, t=14/5

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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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?