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If x^2 + 12x − k = 0, is x = 4?
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05 Jul 2013, 12:41
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Re: If x^2 + 12x − k = 0, is x = 4?
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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.
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Re: If x^2 + 12x − k = 0, is x = 4?
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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!
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Re: If x^2 + 12x − k = 0, is x = 4?
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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.
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Re: If x^2 + 12x − k = 0, is x = 4?
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05 Jul 2013, 13:13



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Re: If x^2 + 12x − k = 0, is x = 4?
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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



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Re: If x^2 + 12x − k = 0, is x = 4?
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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.
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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?
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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.



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Re: If x^2 + 12x − k = 0, is x = 4?
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