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If f(x)=ax²+bx+c, for all x is f(x)<0?

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If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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New post 12 Jun 2017, 01:03
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If f(x)=ax²+bx+c, for all x is f(x)<0?

1) b²-4ac<0
2) a<0

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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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New post 16 Jun 2017, 11:39
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MathRevolution wrote:
If f(x)=ax²+bx+c, for all x is f(x)<0?

1) b²-4ac<0
2) a<0


Please find below theory that should help with this question:
Parabola

A parabola is the graph associated with a quadratic function, i.e. a function of the form \(y=ax^2+bx+c\).


Image

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.

x-intercepts: The x-intercepts, if any, are also called the roots of the function. The x-intercepts are the solutions to the equation \(0=ax^2+bx+c\) and can be calculated by the formula:
\(x_1=\frac{-b-\sqrt{b^2-4ac}}{2a}\) and \(x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}\)

Expression \(b^2-4ac\) is called discriminant:
  • If discriminant is positive parabola has two intercepts with x-axis;
  • If discriminant is negative parabola has no intercepts with x-axis;
  • If discriminant is zero parabola has one intercept with x-axis (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 x-axis. This will happen if discriminant is negative AND the parabola opens downward.

(1) b²-4ac<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/math-coordin ... 87652.html

Hope it helps.
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Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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New post 12 Jun 2017, 09:01
2
MathRevolution wrote:
If f(x)=ax²+bx+c, for all x is f(x)<0?

1) b²-4ac<0
2) a<0


S1 -> b²-4ac<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?  [#permalink]

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New post 12 Jun 2017, 10:41
MathRevolution wrote:
If f(x)=ax²+bx+c, for all x is f(x)<0?

1) b²-4ac<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?  [#permalink]

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New post 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^2-4ac<0, it doesn’t meet with the x-axis, 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?  [#permalink]

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New post 14 Jun 2017, 07:33
If b2-4ac<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?  [#permalink]

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