Polynomial Equation - Roots Formulas

Let the roots of the given equation $$5x^{3} + cx^{2} - 10x + 9 = 0$$ be r, -r and p

r - r + p = $$-\frac{c}{5}$$

p = $$-\frac{c}{5}$$ ...... (1)

$$-r^2-pr+pr=-2$$

$$r^2=2$$ ...... (2)

$$-r^2p=-\frac{9}{5}$$

$$p=\frac{9}{10}$$ ...... (3)

Substituting p in (1), we get

$$\frac{9}{10}=-\frac{c}{5}$$

$$-\frac{9}{2}=c$$

The answer is option A.

Given that -2 is a root of the given cubic equation.

=> Dividing the given equation by (x + 2), Using the Horners method of synthetic division:

coefficient of $$x^2$$ is 1, and coefficient of x is (2r+1)-2 = 2r-1 and the constant term = (4r-1)-2(2r-1) = 1.

=> The quadratic obtained by dividing the cubic = $$x^2+\left(2r-1\right)x+1=0$$, Since, this equation has 2 real roots => Discriminant should be greater than 0

=> $$\left(2r-1\right)^2>4$$ => 2r-1 > 2 or 2r-1 < -2 => r > 3/2 or r < -1/2.

=> Minimum possible non-negative integer value of r is 2.

Given that Amrit gets the sum of the roots correct for the first equation, we can infer that the sum of the roots is 13.
Similarly, from the second equation, we can infer that the product of the roots is 12.
Hence, the equation is $$x^2 - 13x + 12 = 0$$ and the roots are 1 and 12.