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