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In the equation x^2+bx+12=0, x is a variable and b is a
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Updated on: 23 Sep 2018, 19:34
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In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a root of the equation x^2 + bx + 12 = 0. OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/intheequat ... 66784.html
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Originally posted by Baten80 on 21 Feb 2011, 04:38.
Last edited by Bunuel on 23 Sep 2018, 19:34, edited 2 times in total.




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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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21 Feb 2011, 05:40




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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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21 Feb 2011, 05:18
x^2+bx+12=0 a, coefficient of x^2=1 b, coefficient of x=b c, constant in the quadratic polynomial=12 Let \(\alpha,\beta\) be the two factors of the equation. (1) x3 is a factor of x^2+bx+12. \(\alpha=3\) \(\alpha\beta=\frac{c}{a}=\frac{12}{1}\) \(\beta=\frac{12}{3}=4\) \(\alpha+\beta=\frac{b}{a}=\frac{b}{1}\) \(b=7\) Sufficient. (2) 4 is a factor of x^2+bx+12=0 Same as statement 1. \(\beta=4\) \(\alpha\beta=\frac{c}{a}=\frac{12}{1}\) \(\alpha=\frac{12}{4}=3\) \(\alpha+\beta=\frac{b}{a}=\frac{b}{1}\) \(b=7\) Sufficient. Ans: "D"
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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21 Feb 2011, 05:56
Bunuel wrote: Baten80 wrote: In the equation x^2+bx+12=0, is variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0 You don't really need Viete's formula for the roots of a quadratic equation. (1) x3 is a factor of x^2+bx+12 > simply means that x=3 is a root of x^2+bx+12=0 (if x3 is a factor of x^2+bx+12 then x^2+bx+12=(x3)(xk)=0, for some k, so x=3 is on of the roots of the equation) > substitute x=3: 3^2+3b+12=0 > b=7. Sufficient. (2) 4 is a root of x^2+bx+12=0 > the same here, substitute x=4: 4^2+4b+12=0 > b=7. Sufficient. Answer: D. Agree!!! This is much easier and way faster than the Viete's formula (I didn't know that's what it's called). Double thanks!!
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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21 Feb 2011, 05:59
Bunuel wrote: Baten80 wrote: In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a factor of x^2+bx+12=0 You don't really need Viete's formula for the roots of a quadratic equation. (1) x3 is a factor of x^2+bx+12 > simply means that x=3 is a root of x^2+bx+12=0 (if x3 is a factor of x^2+bx+12 then x^2+bx+12=(x3)(xk)=0, for some k, so x=3 is on of the roots of the equation) > substitute x=3: 3^2+3b+12=0 > b=7. Sufficient. (2) 4 is a root of x^2+bx+12=0 > the same here, substitute x=4: 4^2+4b+12=0 > b=7. Sufficient. Answer: D. First post was wrong. This the correct post. But I think answer is same.
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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21 Feb 2011, 06:16
We could also go by the quadratic extensiosn of the equation : Since we know it has to have two rootsr Variables  Statement 1 : X^2  3x4x+12 = 0 x(x3)4(x3) = 0 We have two root  X3 , X4 , So be has to be 3+4 = 7 Sufficient Statement 2: Since X = 4 it has be a root X4, again using the same Expressions as above. We get B=7. Sufficient Ans  D.



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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28 Feb 2011, 01:42
In the equation x^2+bx+12=0, x is a variable and b is a constant. What`s the value of b?
1) x3 is a factor of x^2+bx+12 2) 4 is a root of the equation x^2 + bx + 12=0
Why (1) is sufficient?



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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28 Feb 2011, 01:50
For equation x^2 + bx + 12 = 0, we need to find value of b
1) says x3 is a factor of x^2+bx+12, or x=3 is one of the roots of this quadratic equation, so 3^2+3b+12 = 0 or 21+3b = 0 or b = 7. Sufficient 2) says 4 is one of the roots so 4^2+4b+12 = 0 or 28+4b = 0 or b = 7. Again sufficient, so D
if xa is a factor of any polynomial in x, then at x=a, the value of such a polynomial would be 0 or in other words x = a is one of the roots of such a polynomial



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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15 Dec 2012, 22:40
knew that if x3 is a factor then 3 could be substituted to get x but did not know the meaning of root thanks.. Bunuel, is it helpful to memorize viete's formula?
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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03 Oct 2013, 04:49
D... just find factors, already x3, x4 is another factor to get 12 and 3x4x= 7, this is possible becz x2+ bx +12 = 0, here b = +/, to proof equation only is b= 7
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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24 Oct 2013, 04:53
Bunuel wrote: Baten80 wrote: In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0 You don't really need Viete's formula for the roots of a quadratic equation. (1) x3 is a factor of x^2+bx+12 > simply means that x=3 is a root of x^2+bx+12=0 (if x3 is a factor of x^2+bx+12 then x^2+bx+12=(x3)(xk)=0, for some k, so x=3 is on of the roots of the equation) > substitute x=3: 3^2+3b+12=0 > b=7. Sufficient. (2) 4 is a root of x^2+bx+12=0 > the same here, substitute x=4: 4^2+4b+12=0 > b=7. Sufficient. Answer: D. Im not really following your first point here. How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation?



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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24 Oct 2013, 06:49
waltiebikkiebal wrote: Bunuel wrote: Baten80 wrote: In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0 You don't really need Viete's formula for the roots of a quadratic equation. (1) x3 is a factor of x^2+bx+12 > simply means that x=3 is a root of x^2+bx+12=0 (if x3 is a factor of x^2+bx+12 then x^2+bx+12=(x3)(xk)=0, for some k, so x=3 is on of the roots of the equation) > substitute x=3: 3^2+3b+12=0 > b=7. Sufficient. (2) 4 is a root of x^2+bx+12=0 > the same here, substitute x=4: 4^2+4b+12=0 > b=7. Sufficient. Answer: D. Im not really following your first point here. How do you now that x=3 is a root of the equation? If the equation can be factorized in (X+2)(X+6), isn't three then also a factor from the equation? If (x+2)(x+6)=0, then x3 is NOT a factor of the quadratics. The roots of (x+2)(x+6)=0 are x=3 and x=6. We are told that x3 is a factor of x^2+bx+12, thus (x3)(xk)=0, thus x=3 is one of the roots. Hope it's clear.
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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24 Oct 2013, 09:17
Thanks for your response, but I still don't get why x3 is not a factor. How can we now for sure that it is not a factor?



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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24 Oct 2013, 09:36



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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27 Dec 2014, 13:40
how can root and factor be the same thing? They are not necessarily the same. Am I wrong? Isn't the second option conveying a different meaning?



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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28 Dec 2014, 03:22
A tells x3 a factor means x=3 is a solution so we get value of b...sufficient B tells 4 is a solution ... hence sufficient.



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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10 Nov 2016, 09:22
Bunuel wrote: Baten80 wrote: In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b? (1) x3 is a factor of x^2+bx+12. (2) 4 is a root of x^2+bx+12=0 You don't really need Viete's formula for the roots of a quadratic equation. (1) x3 is a factor of x^2+bx+12 > simply means that x=3 is a root of x^2+bx+12=0 (if x3 is a factor of x^2+bx+12 then x^2+bx+12=(x3)(xk)=0, for some k, so x=3 is on of the roots of the equation) > substitute x=3: 3^2+3b+12=0 > b=7. Sufficient. (2) 4 is a root of x^2+bx+12=0 > the same here, substitute x=4: 4^2+4b+12=0 > b=7. Sufficient. Answer: D. Hello Bunuel, why do you conclude that 4 is a root of the equation if the statement just says it is a factor of (the statement from the page) ... My confusion is that x could be = 12 and 4 would still be a factor of the entire equation. 2) 4 is a factor of \(x^2+bx+12=0\) ... if \(x = 12\), the equation would still be a multiple of 4 and: \((12)^2+12b+12=0\) \(156=12b\) \(13=b\) which is not \(= 7\) and therefore it could be "not sufficient". What am I missing? UPDATE: I looked for the question online and it seems that it has a mistake here, it shouldn't be "factor" ir should say "root". Source: https://books.google.com.mx/books?id=ss ... 3F&f=falsePlease someone correct that, otherwise it is very confusing.
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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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09 Mar 2017, 16:42
Baten80 wrote: In the equation x^2+bx+12=0, x is a variable and b is a constant. What is the value of b?
(1) x3 is a factor of x^2+bx+12. (2) 4 is a factor of x^2+bx+12=0 my answer: D, but i guess i got here by lucky chance... if (x3) is a factor, then the other one will be (x4), and b will be 7. sufficient. i rewrote  if 4 is a factor, then (x4) is a factor, and so is (x3), and I can get to b=7. i guess i was too excited with statement 1...



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Re: In the equation x^2+bx+12=0, x is a variable and b is a
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