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By how much does the larger root of the equation 2x^2+5x = 1

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By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 09:50
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By how much does the larger root of the equation 2x^2+5x = 12 exceed the smaller root?


(A) 5/2
(B) 10/3
(C) 7/2
(D) 14/3
(E) 11/2

[Reveal] Spoiler:
Source: GMATPrep Question Pack 1
Difficulty: Hard
------------
The problem I was having with this question was factoring out

2x^2 + 5x -12 = 0

How do I factor the equation when the x^2 has a coefficient that's not 1 -- in this case 2?

Thanks
[Reveal] Spoiler: OA

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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 10:00
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DelSingh wrote:
By how much does the larger root of the equation 2x^2+5x = 12 exceed the smaller root?


(A) 5/2
(B) 10/3
(C) 7/2
(D) 14/3
(E) 11/2

Source: GMATPrep Question Pack 1
Difficulty: Hard
------------
The problem I was having with this question was factoring out

2x^2 + 5x -12 = 0

How do I factor the equation when the x^2 has a coefficient that's not 1 -- in this case 2?

Thanks


You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 10:00
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DelSingh wrote:
By how much does the larger root of the equation 2x^2+5x = 12 exceed the smaller root?


(A) 5/2
(B) 10/3
(C) 7/2
(D) 14/3
(E) 11/2

Source: GMATPrep Question Pack 1
Difficulty: Hard
------------
The problem I was having with this question was factoring out

2x^2 + 5x -12 = 0

How do I factor the equation when the x^2 has a coefficient that's not 1 -- in this case 2?

Thanks


TO DETERMINE THE ROOTS OF QUADRATIC EQUATION: ax^2+bx+c = 0

formula for root = (-b+\sqrt{(b^2-4ac)})/2a and (-b-\sqrt{(b^2-4ac)})/2a

now in your equation:2x^2+5x-12 = 0
a=2
b=5
c=-12


now when you will plug in the values in the formula

roots come out are = -4 and 3/2

subtracting smaller from bigger will give you 11/2

hope it helps
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 10:27
mau5 wrote:
DelSingh wrote:
By how much does the larger root of the equation 2x^2+5x = 12 exceed the smaller root?


(A) 5/2
(B) 10/3
(C) 7/2
(D) 14/3
(E) 11/2

Source: GMATPrep Question Pack 1
Difficulty: Hard
------------
The problem I was having with this question was factoring out

2x^2 + 5x -12 = 0

How do I factor the equation when the x^2 has a coefficient that's not 1 -- in this case 2?

Thanks


You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.


Note that (a-b)^2 = (a+b)^2-4ab

How are you getting this? I thought (a-b)^2 = a^2 -2ab+ b^2
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 10:30
Expert's post
DelSingh wrote:
mau5 wrote:
DelSingh wrote:
By how much does the larger root of the equation 2x^2+5x = 12 exceed the smaller root?


(A) 5/2
(B) 10/3
(C) 7/2
(D) 14/3
(E) 11/2

Source: GMATPrep Question Pack 1
Difficulty: Hard
------------
The problem I was having with this question was factoring out

2x^2 + 5x -12 = 0

How do I factor the equation when the x^2 has a coefficient that's not 1 -- in this case 2?

Thanks


You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.


Note that (a-b)^2 = (a+b)^2-4ab

How are you getting this? I thought (a-b)^2 = a^2 -2ab+ b^2


(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2

Hope this helps.
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 10:57
mau5 wrote:
You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

How are you getting this? I thought (a-b)^2 = a^2 -2ab+ b^2

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2

Hope this helps.

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2


Sorry for buggin', but I am still curious as to why you chose to manipulate (a-b)^2 into (a+b)^2-4ab when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know :)
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 11:06
Expert's post
DelSingh wrote:
mau5 wrote:
You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

How are you getting this? I thought (a-b)^2 = a^2 -2ab+ b^2

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2

Hope this helps.

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2


Sorry for buggin', but I am still curious as to why you chose to manipulate (a-b)^2 into (a+b)^2-4ab when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know :)


I chose this method only in this context . The question was asking for the difference of roots.
.
Now, we alrady know the sum and the product of the 2 roots. The formula which I have used is just to get the difference of the 2.

By the way , it might be a handy formula to remember.

Hope this helps.
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 04 Aug 2013, 11:07
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DelSingh wrote:
mau5 wrote:
You don't need to factor out anything for this question. Note that (a-b)^2 = (a+b)^2-4ab

Now, the sum of the roots :\frac{-5}{2} and product of the roots :\frac{-12}{2}

Thus, (a-b)^2 = \frac{25}{4}+4*6 = \frac{121}{4}

Thus, (a-b) = \frac{11}{2}

E.

Note that (a-b)^2 = (a+b)^2-4ab

How are you getting this? I thought (a-b)^2 = a^2 -2ab+ b^2

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2

Hope this helps.

(a+b)^2-4ab = a^2+b^2+2ab-4ab = a^2 -2ab+ b^2


Sorry for buggin', but I am still curious as to why you chose to manipulate (a-b)^2 into (a+b)^2-4ab when you encountered this problem? Is there some sort of method/property that comes to mind? The study guides I am using doesn't really show this, but I would love to know :)


for a quadratic equation AX^2+BX+C = 0
SUM OF ROOTS = -B/A
PRODUCT OF ROOTS = C/A
let a AND b be the roots of equation
then a*b = C/A
a + b = -B/A

now as we have to calculate difference of roots (a-b)
we can use directly the formula (a-b)^2 = (a+b)^2 - 4ab...now simply you have to plug in the values..

hope it helps
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By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 31 Oct 2013, 00:25
Re-writing the equation as follows:

2x^2 + 5x - 12 =0

The two factors of -24 (-12 x 2) are 8 & -3

So,2x^2 + 8x - 3x - 12 = 0

2x (x +4) -3 (x+4) = 0

(2x -3) (x+4) = 0

So x = \frac{3}{2} or x = -4
Distance between 3/2 & -4 =

\frac{3}{2} - (-4) = \frac{11}{2} (Answer = E)
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Last edited by PareshGmat on 03 Nov 2014, 00:23, edited 1 time in total.
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink] New post 02 Nov 2014, 22:56
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Re: By how much does the larger root of the equation 2x^2+5x = 1   [#permalink] 02 Nov 2014, 22:56
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