If f(x) = ax^2 + bx + c, for all x is f(x)

Question

If  f(x) = ax^2 + bx + c , for all x is  f(x) < 0 ?

(1)  b^2 - 4ac < 0 
(2)  a < 0

Bunuel · Accepted Answer

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 . https://gmatclub.com/forum/download/file.php?id=10339 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.

akshayk · Answer

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.

Mo2men · Answer

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

MathRevolution · Answer

==> 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

vivek2k70 · Answer

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

ohhno · Answer

This is a Data Sufficiency question. We need to answer:

Quote:

Is f(x)<0f(x)<0 for every value of xx?

We have:
f(x)=ax2+bx+cf(x)=ax^2+bx+c
For a quadratic to be always negative, two things must happen:

It must open downward → a<0a<0
It must never touch or cross the x-axis → discriminant b2−4ac<0b^2-4ac<0

So we actually need both conditions.
Statement (1)
b2−4ac<0b^2-4ac<0
This tells us the quadratic has no real roots, but it doesn't tell us whether it is always positive or always negative.
For example:
f(x)=x2+1f(x)=x^2+1
has discriminant 0−4=−4<00-4=-4<0, but f(x)>0f(x)>0 for all xx.
❌ Not sufficient.

Statement (2)
a<0a<0
This tells us the parabola opens downward, but it could still cross the x-axis.
Example:
f(x)=−x2+1f(x)=-x^2+1
Here a<0a<0, but f(0)=1>0f(0)=1>0.
❌ Not sufficient.

Together
We have:
a<0a<0
and
b2−4ac<0b^2-4ac<0
genui{"graphable_function_v2_learning_block_parameterized":{"expressions":[{"latex":"y=ax^2+bx+c"},{"latex":"a<0"},{"latex":"b^2-4ac<0"}]}}
A downward-opening quadratic with no real roots must lie entirely below the x-axis.
Therefore:
f(x)<0 for all x\boxed{f(x)<0\text{ for all }x}
✅ Answer: (C) — Both statements together are sufficient, but neither alone is sufficient.

MathRevolution

If \(f(x) = ax^2 + bx + c\), for all x is \(f(x) < 0\)?

(1) \(b^2 - 4ac < 0\)
(2) \(a < 0\)

If f(x) = ax^2 + bx + c, for all x is f(x) < 0? (1) b^2 - 4ac < 0 (2)

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