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By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
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04 Aug 2013, 10:50
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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 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
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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 \((ab)^2 = (a+b)^24ab\) Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\) Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\) Thus, \((ab) = \frac{11}{2}\) 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 = \((b+\sqrt{(b^24ac)})/2a\) and \((b\sqrt{(b^24ac)})/2a\) now in your equation:\(2x^2+5x12 = 0\) \(a=2 b=5 c=12\) 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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04 Aug 2013, 11:27
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 \((ab)^2 = (a+b)^24ab\) Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\) Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\) Thus, \((ab) = \frac{11}{2}\) E. Note that \((ab)^2 = (a+b)^24ab\) How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\)
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
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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 \((ab)^2 = (a+b)^24ab\) Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\) Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\) Thus, \((ab) = \frac{11}{2}\) E. Note that \((ab)^2 = (a+b)^24ab\) How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\) \((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\) 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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04 Aug 2013, 11:57
mau5 wrote: You don't need to factor out anything for this question. Note that \((ab)^2 = (a+b)^24ab\)
Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\)
Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\)
Thus, \((ab) = \frac{11}{2}\)
E.
Note that \((ab)^2 = (a+b)^24ab\)
How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\)
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\)
Hope this helps.
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\) Sorry for buggin', but I am still curious as to why you chose to manipulate \((ab)^2 into (a+b)^24ab\) 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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04 Aug 2013, 12:06
DelSingh wrote: mau5 wrote: You don't need to factor out anything for this question. Note that \((ab)^2 = (a+b)^24ab\)
Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\)
Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\)
Thus, \((ab) = \frac{11}{2}\)
E.
Note that \((ab)^2 = (a+b)^24ab\)
How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\)
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\)
Hope this helps.
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\) Sorry for buggin', but I am still curious as to why you chose to manipulate \((ab)^2 into (a+b)^24ab\) 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 \((ab)^2 = (a+b)^24ab\)
Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\)
Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\)
Thus, \((ab) = \frac{11}{2}\)
E.
Note that \((ab)^2 = (a+b)^24ab\)
How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\)
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\)
Hope this helps.
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\) Sorry for buggin', but I am still curious as to why you chose to manipulate \((ab)^2 into (a+b)^24ab\) 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 (ab) we can use directly the formula (ab)^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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Rewriting 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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31 Mar 2015, 05:20
DelSingh wrote: mau5 wrote: You don't need to factor out anything for this question. Note that \((ab)^2 = (a+b)^24ab\)
Now, the sum of the roots :\(\frac{5}{2}\) and product of the roots :\(\frac{12}{2}\)
Thus, \((ab)^2 = \frac{25}{4}+4*6 = \frac{121}{4}\)
Thus, \((ab) = \frac{11}{2}\)
E.
Note that \((ab)^2 = (a+b)^24ab\)
How are you getting this? I thought \((ab)^2 = a^2 2ab+ b^2\)
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\)
Hope this helps.
\((a+b)^24ab = a^2+b^2+2ab4ab = a^2 2ab+ b^2\) Sorry for buggin', but I am still curious as to why you chose to manipulate \((ab)^2 into (a+b)^24ab\) 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 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=(bsquare root of Discriminant)/2a. So you can find both roots and solve the problem. For example in our case Discriminant=254*(12)*2=121. Hence x1=(5+11)/4=1.5 and x2=(511)/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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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 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).  Optimus Prep's GMAT On Demand course for only $299 covers all verbal and quant. concepts in detail. Visit the following link to get your 7 days free trial account: http://www.optimusprep.com/gmatondemandcourse
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Re: By how much does the larger root of the equation 2x^2+5x = 1 [#permalink]
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02 Apr 2015, 00:37
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 FOILing 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: GMAT assassins aren't born, they're made, Rich
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