algebra 2
posted by lizzie .
I don't know how to do this.
Which describes the number and type of roots of the equation x^3  4x^2 + 50x + 7 = 0?
a. 1 positive, 2 negative
b. 2 positive, 1 negative
c. 3 negative
d. 3 positive.

Let X1, X2, X3 are roots of the equation,
then
X1+X2+X3=4
X1*X2+X1*X3+X2*X3=50
X1*X2*X3=7
We can choose only b. 
I don't know of any easy way to do this, so I used a website to provide a numerical solution.
None of the choic es are correct. Two of the roots are complex. The other is negative (0.138414..)
Make sure you copied the problem correctly. 
There are two changes in the sign of the coefficients, so we have either 2 positive roots, or none.
Also, by negating the oddpowered terms, there is one change of sign, so we have 1 negative root. This invalidates choices a, c and d.
Without actually solving the equation, the choice is either
(b), or
"none of the above", i.e. 2 complex and one negative root (as confirmed by drwls's solution).
Another way to see that there are 2 complex roots without solving the equation is to look at
f'(x)=0, or
3x^28x+50=0
which does not have real roots implying no maximum nor minimum. Therefore f(x) is monotonically increasing, and therefore has only one real root.
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