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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]

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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]

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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]

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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]

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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]

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New post 04 Aug 2013, 11:30
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]

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New post 04 Aug 2013, 11:57
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]

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New post 04 Aug 2013, 12:06
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]

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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]

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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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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]

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New post 31 Mar 2015, 05:20
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 :)

Hi Delsingh!
I looked through the discussion and decided to help you if you still need help :-D :-D
So have ever heard of Discriminant? This variable allows us to find roots of the equation without factoring. If you have a quadratic equation like a(x^2)+bx+c=0 then you can find Discriminant and roots. Formula for Discriminant is (b^2)-4ac. Formula for roots is x1=(-b+square root of Discriminant)/2a and for x2=(-b-square root of Discriminant)/2a. So you can find both roots and solve the problem. For example in our case Discriminant=25-4*(-12)*2=121. Hence x1=(-5+11)/4=1.5 and x2=(-5-11)/4=-4 Hope it is clear
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]

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New post 31 Mar 2015, 11:54
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

[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


Roots are [-5 + sqrt(25 + 96)]/4 OR [-5 - sqrt(25 + 96)]/4
= 1.5 OR -4
Hence larger root 1.5 is 1.5 - (-4) = 5.5 = 11/2 greater than smaller root (-4).
Hence option (E).

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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]

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I do not think this question is from gmatprep.
pls show the picture
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]

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Hi All,

It looks like a number of the explanations take a more complex approach than what is needed. This question is based on FOIL-ing and Factoring rules; even though it looks a little "tough", the same rules still apply...

We're given 2X^2 + 5X = 12

We can rewrite that as....

2X^2 + 5X - 12 = 0

Now let's break this into it's two 'pieces'....

(X _ _ )(2X _ _ )

Now let's look at the '-12'....

This means that the two numbers could be....
1 and 12
2 and 6
3 and 4

And one number is + and the other is -

Since the middle term of the Quadratic is "5X", we need to 'play around' a bit with the possibilities....

1 and 12 are too far 'apart'
2 and 6 are both even, so we won't end up with 5X (since 5 is odd)

That just leaves us with 3 and 4....
(X + 4)(2X - 3) = 0

Now we can solve the Quadratic...

X = -4, +3/2

The prompt asks for the difference in the solutions...
(3/2) - (-4) = 11/2

Final Answer:
[Reveal] Spoiler:
E


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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]

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New post 16 Nov 2017, 18:08
PareshGmat wrote:
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)


Solved in same way as above one.I personally feel that's the best approach rather than remembering any formula.

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Re: By how much does the larger root of the equation 2x^2+5x = 1   [#permalink] 16 Nov 2017, 18:08
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