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Math Revolution GMAT Instructor
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If f(x)=ax²+bx+c, for all x is f(x)<0?
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12 Jun 2017, 01:03
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43% (01:17) correct 57% (01:25) wrong based on 81 sessions
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If f(x)=ax²+bx+c, for all x is f(x)<0? 1) b²4ac<0 2) a<0
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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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16 Jun 2017, 11:39
MathRevolution wrote: If f(x)=ax²+bx+c, for all x is f(x)<0?
1) b²4ac<0 2) a<0 Please find below theory that should help with this question: ParabolaA parabola is the graph associated with a quadratic function, i.e. a function of the form \(y=ax^2+bx+c\). The general or standard form of a quadratic function is \(y =ax^2+bx+c\), or in function form, \(f(x)=ax^2+bx+c\), where \(x\) is the independent variable, \(y\) is the dependent variable, and \(a\), \(b\), and \(c\) are constants.  If \(a\) is positive, the parabola opens upward, if negative, the parabola opens downward.
xintercepts: The xintercepts, if any, are also called the roots of the function. The xintercepts are the solutions to the equation \(0=ax^2+bx+c\) and can be calculated by the formula: \(x_1=\frac{b\sqrt{b^24ac}}{2a}\) and \(x_2=\frac{b+\sqrt{b^24ac}}{2a}\) Expression \(b^24ac\) is called discriminant:  If discriminant is positive parabola has two intercepts with xaxis;
 If discriminant is negative parabola has no intercepts with xaxis;
 If discriminant is zero parabola has one intercept with xaxis (tangent point).
BACK TO THE QUESTION: If f(x)=ax²+bx+c, for all x is f(x)<0? According to the theory presented above, the question asks whether entire parabola is below xaxis. This will happen if discriminant is negative AND the parabola opens downward. (1) b²4ac<0 > the discriminant is negative. Not sufficient. (2) a<0 > the parabola opens downward. Not sufficient. (1)+(2) Both conditions satisfied. Sufficient. Answer: C. For more check here: https://gmatclub.com/forum/mathcoordin ... 87652.htmlHope it helps.
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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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12 Jun 2017, 09:01
MathRevolution wrote: If f(x)=ax²+bx+c, for all x is f(x)<0?
1) b²4ac<0 2) a<0 S1 > b²4ac<0 => a and c are the same sign, can be negative and can be positive. If a & c are positive, f(x) > 0, but if a & c are negative, f(x)<0 Insufficient. S2 > a<0 No information about c => insufficient. S1+S2 a & c are the same sign and a<0 which implies c<0 Therefore, f(x)<0 for any value of x. C is the answer IMO.
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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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12 Jun 2017, 10:41
MathRevolution wrote: If f(x)=ax²+bx+c, for all x is f(x)<0?
1) b²4ac<0 2) a<0 Dear GMATPrepNow, Can you please share your thoughts about the math behind this question? As I know, when the discriminate is negative, there is no real solution exits. Thanks in advance



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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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14 Jun 2017, 01:25
==> In the original condition, there are 3 variables (a, b, c) and in order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), if discriminant =b^24ac<0, it doesn’t meet with the xaxis, and if a<0, you always get f(x)<0, hence yes, it is sufficient. Therefore, the answer is C. Answer: C
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If f(x)=ax²+bx+c, for all x is f(x)<0?
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14 Jun 2017, 07:33
If b24ac<0 then it is complex number. Also we don't know the value of b here. Can someone explain in detail the logic behind in this question



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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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21 Mar 2019, 16:47
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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?
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