# Math Test Help

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• Feb 14th 2010, 04:26 PM
sleigh
Math Test Help
Hi. I was taking a practice test for a Math Contest and I can not figure out what the answer to one question is. Here is the question:

The function $\displaystyle f(x)=x^3+7x^2-6x+72$has roots $\displaystyle r1 , r2 , r3$ (i.e. $\displaystyle f(r1)=f(r2)=f(r3)=0$). Find $\displaystyle r1+r2+r3$.

A. 7
B. -7
C. 0
D. -5
E. None of these.

Correct answer: B

I used the zeros function on the TI 89, and it said that the only zero of the function was -8.65..., so I put E, but I don't see why this is wrong. Could someone explain this question to me? Thanks!
• Feb 14th 2010, 05:30 PM
Archie Meade
Quote:

Originally Posted by sleigh
Hi. I was taking a practice test for a Math Contest and I can not figure out what the answer to one question is. Here is the question:

The function $\displaystyle f(x)=x^3+7x^2-6x+72$has roots $\displaystyle r1 , r2 , r3$ (i.e. $\displaystyle f(r1)=f(r2)=f(r3)=0$). Find $\displaystyle r1+r2+r3$.

A. 7
B. -7
C. 0
D. -5
E. None of these.

Correct answer: B

I used the zeros function on the TI 89, and it said that the only zero of the function was -8.65..., so I put E, but I don't see why this is wrong. Could someone explain this question to me? Thanks!

Hi sleigh,

the answer would be B if there was a -72 at the end.
The roots of that are -6, -4 and 3.
• Feb 14th 2010, 05:39 PM
TKHunny
This is an eyeball problem. Every moment you spent on your calculator was wasted. If this is USAD, then you have 30 minutes to solve 35 problems. You do not have time to waste. Just mark the correct answer and move on.

Try this (but not on the exam)

A polynomial has three roots: a, b, c

The polynomial looks like this (x-a)(x-b)(x-c)

If we expand this polynomial, we get this: x^3 - (a+b+c)x^2 + (ab + ac + bc)x - abc

If they ask for abc, you simply read off the constant term and change the sign.

If they ask for a+b+c, you simply read off the coefficient on the term of second highest degree and change the sign.

In this case, you mark -7 and move on.

Your TI89 choked on the two Complex Solutions. You could have used the one Real solution to reduce the cubic to a quadratic. Then, the other two would become apparent. Still, while you were doing this, you could have solved three other problems. Don't throw time out the window.
• Feb 14th 2010, 05:43 PM
TKHunny
I am delighted by the wonderful lack of significance of the constant term in this problem. Archie Mead showed an excellent integer solution. Change the constant term to $\displaystyle \sqrt{3}$ and the answer remains "-7".

It's a beautiful thing. (Cool)
• Feb 14th 2010, 06:08 PM
sleigh
Thank you everyone for you help! I try not to use my calculator when taking tests, but I didn't know any other way to solve this one. Thanks again!