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 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.
To determine the number and type of roots for the equation x^3 - 4x^2 + 50x + 7 = 0, you can use an algebraic method called the Descartes' rule of signs.
1. Begin by counting the sign changes in the equation. In this case, we start with a positive coefficient of x^3 and then alternate signs between subsequent terms (from -4x^2 to +50x), resulting in two sign changes.
2. Next, compute the sign changes in the equation when substituting -x for x. By substituting -x into the equation, we have (-x)^3 - 4(-x)^2 + 50(-x) + 7 = -x^3 - 4x^2 - 50x + 7 = 0. Here, we begin with a negative coefficient of x^3 and again alternate signs between subsequent terms (from -4x^2 to -50x), giving us two more sign changes.
3. The Descartes' rule of signs states that the number of positive roots of the equation is equal to the number of sign changes or less by an even number. Hence, in this case, we can conclude that the equation has either three positive roots or one positive root.
4. Similarly, the Descartes' rule of signs also tells us that the number of negative roots of the equation is equal to the number of sign changes or less by an even number. So, we can conclude that there are either two negative roots or no negative roots.
Based on the given options, none of them match the possibilities implied by the Descartes' rule of signs. Therefore, none of the given options describe the correct number and type of roots for this equation.