Posted by **lizzie** on Sunday, August 14, 2011 at 10:03am.

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.

- algebra 2 -
**Anonymous**, Sunday, August 14, 2011 at 10:54am
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.

- algebra 2 -
**drwls**, Sunday, August 14, 2011 at 10:58am
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.

- algebra 2 -
**MathMate**, Sunday, August 14, 2011 at 11:35am
There are two changes in the sign of the coefficients, so we have either 2 positive roots, *or none*.

Also, by negating the odd-powered 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^2-8x+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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