Author 
Message 
TAGS:

Hide Tags

Retired Moderator
Joined: 27 Aug 2012
Posts: 1044

If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
05 Jul 2013, 12:41
Question Stats:
53% (02:01) correct 47% (02:07) wrong based on 235 sessions
HideShow timer Statistics
If x^2 + 12x − k = 0, is x = 4? (1) (x + 16) is a factor of x^2 + 12x − k = 0, where k is a constant, and x is a variable. (2) x ≠ −16
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 60625

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
05 Jul 2013, 12:54
If x^2+12x−k=0, is x=4?(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation. Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\). So, we have that x is either 16 or 4. Not sufficient. (2) x≠−16. Clearly insufficient. (1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient. Answer: C. Hope it's clear.
_________________




Retired Moderator
Joined: 27 Aug 2012
Posts: 1044

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
05 Jul 2013, 13:01
Bunuel wrote: If x^2+12x−k=0, is x=4?
(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation.
Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\).
So, we have that x is either 16 or 4. Not sufficient.
(2) x≠−16. Clearly insufficient.
(1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient.
Answer: C.
Hope it's clear. (1)+(2)=> 1 says x=16 and 2 says x is NOT 16...So,isn't it contradicting hence Insufficient..? I'm having confusion at this point...Please help me understand!
_________________



Math Expert
Joined: 02 Sep 2009
Posts: 60625

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
05 Jul 2013, 13:04
debayan222 wrote: Bunuel wrote: If x^2+12x−k=0, is x=4?
(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation.
Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\).
So, we have that x is either 16 or 4. Not sufficient.
(2) x≠−16. Clearly insufficient.
(1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient.
Answer: C.
Hope it's clear. (1)+(2)=> 1 says x=16 and 2 says x is NOT 16...So,isn't it contradicting hence Insufficient..? I'm having confusion at this point...Please help me understand! (1) says that x=16 OR x=4. The equation is \(x^2+12x64=0\) (k=64) > x=16 OR x=4.
_________________



Retired Moderator
Joined: 27 Aug 2012
Posts: 1044

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
05 Jul 2013, 13:13
Got it! I was doing the mistake by considering the Stat.1 ONLY not focusing on the solns of the eqn in 1...! Thanks for clarifying.
_________________



Intern
Joined: 27 Jun 2013
Posts: 21
Location: Russian Federation
GMAT 1: 620 Q50 V27 GMAT 2: 730 Q50 V39

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
22 Aug 2013, 08:37
Bunuel wrote: If x^2+12x−k=0, is x=4?
(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation.
Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\).
So, we have that x is either 16 or 4. Not sufficient.
(2) x≠−16. Clearly insufficient.
(1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient.
Answer: C.
Hope it's clear. Why A is not sufficient? X cannot be 16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan



Math Expert
Joined: 02 Sep 2009
Posts: 60625

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
22 Aug 2013, 09:46
Dmitriy wrote: Bunuel wrote: If x^2+12x−k=0, is x=4?
(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation.
Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\).
So, we have that x is either 16 or 4. Not sufficient.
(2) x≠−16. Clearly insufficient.
(1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient.
Answer: C.
Hope it's clear. Why A is not sufficient? X cannot be 16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x4) are factors of x^2+12x−64=0 > (x+16)(x4)=0. Hope this helps.
_________________



Intern
Joined: 27 Jun 2013
Posts: 21
Location: Russian Federation
GMAT 1: 620 Q50 V27 GMAT 2: 730 Q50 V39

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
28 Aug 2013, 22:40
Bunuel wrote: Dmitriy wrote: Bunuel wrote: If x^2+12x−k=0, is x=4?
(1) (x+16) is a factor of x^2+12x−k=0, where k is a constant, and x is a variable. This implies that we can factor out \(x+16\) from \(x^2+12xk=0\), so we would have \((x+16)*(something)=0\). Thus x=16 is one of the roots of the given quadratic equation.
Viete's theorem states that for the roots \(x_1\) and \(x_2\) of a quadratic equation \(ax^2+bx+c=0\):
\(x_1+x_2=\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).
Thus according to the above \(x_1+x_2=16+x_2=\frac{12}{1}\) > \(x_2=4\).
So, we have that x is either 16 or 4. Not sufficient.
(2) x≠−16. Clearly insufficient.
(1)+(2) Since (2) says that x is NOT 16, then x=4. Sufficient.
Answer: C.
Hope it's clear. Why A is not sufficient? X cannot be 16, as in this case X+16 =0. As I know 0 cannot be a factor or the integer. Pls. explan Factor of a number and factor of an expression are two different things. For example, both (x+16) and (x4) are factors of x^2+12x−64=0 > (x+16)(x4)=0. Hope this helps. Thanks. I didnt know about factors of an expression.



NonHuman User
Joined: 09 Sep 2013
Posts: 13999

Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
Show Tags
15 Jul 2019, 17:41
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If x^2 + 12x − k = 0, is x = 4?
[#permalink]
15 Jul 2019, 17:41






