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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]  05 Jul 2013, 11:41
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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]  05 Jul 2013, 11:54
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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.

Hope it's clear.
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Re: If x^2+12x−k=0, is x=4? [#permalink]  05 Jul 2013, 12:01
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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.

Hope it's clear.

(1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..?

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Re: If x^2+12x−k=0, is x=4? [#permalink]  05 Jul 2013, 12:04
Expert's post
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.

Hope it's clear.

(1)+(2)=> 1 says x=-16 and 2 says x is NOT -16...So,isn't it contradicting hence Insufficient..?

(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]  05 Jul 2013, 12:13
Expert's post
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]  22 Aug 2013, 07:37
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.

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]  22 Aug 2013, 08:46
Expert's post
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.

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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Re: If x^2+12x−k=0, is x=4? [#permalink]  28 Aug 2013, 21:40
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.

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.
Re: If x^2+12x−k=0, is x=4?   [#permalink] 28 Aug 2013, 21:40
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