Multiple choice question about the roots of a cubic equation

Question

I am in class $7$th and studying high school maths. I encountered the following question. How can I solve it?

Well I have eliminated option A and B by using the concept that sum of roots in cubic equation is equal to $b/a$ and product of the roots is $d/a$. Here is my working:

$$x^2-ax^2-bx+c=0$$ Let the roots be $j$, $k$, and $l$. Then, \begin{align} j + k + l &= a \\ jkl &= -c \\ jk + kl + lj &= -b \end{align}

Answer 1

x1 + x2 + x3 =-a and x1x2x3 =-c

This means that

x1 + x2 + x3 >0 x1x2x3 <0

This is possible when two roots are greater than 0, and one less than 0

Answer 2

Since OP indicates that they are in a high school course, it may be that the question only intended that the students use Descartes' Rule of Signs. If we include the signs of the coefficients described in the problem statement, we have $$ s(x) \ = \ x^3 - ax^2 - bx + c $$ and $$ s(-x) \ = \ (-x)^3 - a(-x)^2 - b(-x) + c \ = \ -x^3 - ax^2 + bx + c \ \ . $$

If we scan along the coefficients in each expression, we note that the sign of the coefficients changes twice in $ \ s(x) \ $ (positive to negative and back again), while the sign only changes once in $ \ s(-x) \ $ (negative to positive). The Rule tells us that there are either two or zero positive zeroes for this polynomial and one negative zero.

The "hedging" in Descartes' Rule is due to the fact that these sign changes cannot tell us about irreducible quadratic factors of the polynomial which would have complex zeroes. However, a further condition in the problem is that the zeroes are all integers: this is stronger than strictly necessary, as it will suffice to make our choice to know that all of the zeroes are real. Consequently, we now know that the polynomial as specified has two positive and one negative real zero; this corresponds to choice $ \ \mathbf{(D)} \ \ . $

Answer 3

Suppose the solutions to the cubic $x^3+ax^2+bx+c$ are $r_1$, $r_2$, $r_3$. The the factored form of the equation is

$$(x-r_1)(x-r_2)(x-r_3)=0$$

Expanding the equation gives

$$x^3-(r_1+r_2+r_3)x^2+(r_1r_2+r_1r_3+r_2r_3)x-r_1r_2r_3=0$$

Since it’s given that $a,b\lt0$ and $c\gt0$,

$$r_1+r_2+r_3\gt0$$ $$r_1r_2+r_1r_3+r_2r_3\gt0$$ $$r_1r_2r_3\lt0$$

The first and third conditions are enough to determine the answer. For the product of the roots to be negative, either one root is negative or all three roots are negative. If the sum of the roots is positive, then all the roots can’t be negative.

Therefore, one root is negative, the other two are positive. (D)

Note: relabeling coefficients so that a, b, c are positive is fine, but you have to be careful to keep track of that change. It can cause confusion. In this case, the coefficients were only needed to set up the inequalities. They constraint didn’t really matter until the end.