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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 41% (01:14) correct 59% (01:21) wrong based on 94 sessions

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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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Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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1
6
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 $$y=ax^2+bx+c$$. 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.

For more check here: https://gmatclub.com/forum/math-coordin ... 87652.html

Hope it helps.
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Senior Manager  G
Joined: 06 Jul 2016
Posts: 356
Location: Singapore
Concentration: Strategy, Finance
Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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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

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.
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Joined: 26 Mar 2013
Posts: 2345
Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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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

Dear GMATPrepNow,

As I know, when the discriminate is negative, there is no real solution exits.

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8261
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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==> 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.

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Joined: 25 Dec 2016
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If f(x)=ax²+bx+c, for all x is f(x)<0?  [#permalink]

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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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