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If x^2+12x−k=0, is x=4?

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If x^2+12x−k=0, is x=4? [#permalink]

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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


P.S: Bunuel- Need your help Sir!
[Reveal] Spoiler: OA

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

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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 \(x^2+12x-k=0\), 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.
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Re: If x^2+12x−k=0, is x=4? [#permalink]

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New post 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 \(x^2+12x-k=0\), 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!
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Re: If x^2+12x−k=0, is x=4? [#permalink]

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New post 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 \(x^2+12x-k=0\), 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 \(x^2+12x-64=0\) (k=64) --> x=-16 OR x=4.
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Re: If x^2+12x−k=0, is x=4? [#permalink]

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New post 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.
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Re: If x^2+12x−k=0, is x=4? [#permalink]

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New post 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 \(x^2+12x-k=0\), 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

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

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New post 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 \(x^2+12x-k=0\), 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 (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0.

Hope this helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 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 \(x^2+12x-k=0\), 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 (x-4) are factors of x^2+12x−64=0 --> (x+16)(x-4)=0.

Hope this helps.

Thanks. I didnt know about factors of an expression.

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

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

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Re: If x^2+12x−k=0, is x=4?   [#permalink] 08 Sep 2016, 06:44
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