Quadratic Fun!

Algebra Level 3

f ( x ) = a x 2 + b x + c f(x)=ax^2+bx+c . a , b , c a,b,c are real numbers. Given that c ( a + b + c ) < 0 c(a+b+c)<0 , then what can we say about b 2 − 4 a c b^2-4ac ?

b 2 − 4 a c = 0 b^2 - 4ac = 0 b 2 − 4 a c < 0 b^2 - 4ac < 0 No conclusion can be drawn b 2 − 4 a c > 0 b^2 - 4ac > 0

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Dhiraj Agarwalla by dhiraj agarwalla , 23, India

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6 solutions

Dhiraj Agarwalla
Sep 28, 2014

c(a+b+c) is same as f(0) f(1). As f(0) f(1) is less than 0, we can say that f(0)and f(1) are of opposite signs. Therefore the graph of f(x) will cut the x-axis and therefore the discriminant of the quadratic equation has to be greater than 0.

40 Helpful 2 Interesting 2 Brilliant 0 Confused

Completely and unnecessarily misleading to provide a graphic where the parabola does not cut the axis. If you wanted to provide a graphic at all, then show three parabola for the three cases, instead of suggesting the wrong answer with this one.

Andrew Rakich - 6 years, 8 months ago

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I was misleaded by teh graph and also because that was no "<0" in my sentence when I got the e-mail. I did not bother to read again , just thought it was useless information when I got this page.

Look at the e-mail:

f(x) = ax2 + bx + c. a, b, c are real numbers. Given that c(a + b + c) , then what can we say about b2 - 4ac?

David Bronstein - 6 years, 8 months ago

ya I gave ans through graph only it became easy for me

Ankit Nigam - 6 years, 8 months ago

Interesting question. I've updated the answer into Multiple choice, so that it is easier to understand what you are asking for.

Note: Technically it is still true that b 2 − 4 a c > − 10 b^2 - 4ac > - 10 , which is why I didn't like the original phrasing.

Calvin Lin Staff - 6 years, 8 months ago

I just latexed some of your problems @dhiraj agarwalla . You should also use latex in your problems wherever required. If you don't know latex, you can always learn it here . Thanks.

Satvik Golechha - 6 years, 8 months ago

actually I agree with the point of misleading graphic

chirag shetty - 6 years, 8 months ago

everyone please the answer is not correct i will prove it C(A+B+C) < 0

THEN A+B+C < 0

B < -A-C

B^2 < A^2 +2AC + C^2

BY adding -4AC to both sides

B^2 - 4AC < A^2 - 4AC +C^2

B^2 - 4AC < ( A-C )^2

SO IF A =C THEN B^2 -4AC <0

IF A IS NOT = C THEN B^2 -4AC < POSITIVE NUMBER THAT CAN MAKE B^2 -4AC WHETHER IT CAN BE POSITIVE OR EQUAL ZERO OR LESS THAN ZERO

TAKE A GOOD LOOK PLEASE THE REAL ANSWER IS ( NO CONCLUSION )

Omar Khedr - 6 years, 6 months ago

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Omar there's a little goof up in your third step where you multiply -1 two times(square it ) and don't change the sign of the inequality. Hence, the answer remains correct.

Kunal Verma - 6 years, 4 months ago

In the first step, when you divide both side c, you are taking c as a positive number what makes (a+b+c)<0, but if c is negative it means that (a+b+c)>0

Guilherme Ferreira Carvalho - 5 years, 5 months ago
Sukhveer Sahota
Jul 22, 2017

We manipulate the given inequality as follows
c(a + b + c) < 0
ac + bc + c^2 < 0
ac < -c^2 - bc
-4ac > 4c^2 + 4bc
b^2 - 4ac > b^2 + 4c^2 + 4bc
b^2 - 4ac > (b+2c)^2


Knowing that b and c are real numbers, the expression (b+2c)^2 will always be greater than or equal to 0. It follows that b^2 - 4ac is also greater than 0.

16 Helpful 2 Interesting 6 Brilliant 0 Confused
Christian Daang
Oct 18, 2014

we can say:

Let:

c < (a,b) where (a,b) are Positive #'s ...

Then,

c(a+b+c) < 0

That results negative,,

then,

a and b are both positive numbers and c is a negative number,,

then...

b^2 - 4ac

= (Positive) - 4(Positive)(Negative)

= Positive,

So..

The discriminant will be Positive and > 0 ... :D

6 Helpful 0 Interesting 0 Brilliant 1 Confused
Lucas Chen
Nov 6, 2014

c(a+b+c)<0, so a+b+c<0. This made me to lead that a,b,c were all negative, a was pos and b and c were negative... etc. Then notice how everytime this leads to b^2-4ac being positive. Sorry for the childish solution.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

As c x(a+b+c) is less than 0.So c is negative.For negative value of c the negative sign of 4 x a x c becomes positive.Thus b square - 4 x a x c is positive.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Richard Levine
Oct 5, 2014

The quadratic formula states that x = (-b +- squareroot(b^2 - 4ac))/2a. If b^2 - 4ac were less than 0, the solution would yield an imaginary number (no real number solution). If b^2 - 4ac were 0, then according to the formula -b/2a would have to be 0, so b would be 0. But then -4ac is 0, and that means a or c is 0. If a is 0 and b is 0, then f(x)=c, and that would produce a straight line. If c is 0, then c(a+b+c) is not less than 0. Therefore, the choice is b^2 - 4ac is greater than 0.

0 Helpful 0 Interesting 0 Brilliant 1 Confused

A great "but" to this problem's title is that is not how much do you know about quadratic formula, but how much are you able to recall and connect with Bolzano's Theorem, f continuous and different sign in extremes of an interval, then there must exist at leas one solution, leading to that it must be two, therefore discriminant > 0

Jose Torres Zapata - 6 years, 8 months ago

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