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Re: If x^2+12x−k=0, is x=4? [#permalink]
05 Jul 2013, 11:54
2
This post received KUDOS
Expert's post
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This post was BOOKMARKED
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.
Re: If x^2+12x−k=0, is x=4? [#permalink]
05 Jul 2013, 12:01
Expert's post
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! _________________
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.
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. _________________
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.
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
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.
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.
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.
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.
Re: If x^2+12x−k=0, is x=4? [#permalink]
09 Sep 2014, 06:28
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