Question

Which correctly describes the roots of the following cubic equation x^3-5x^2+3x+9=0? A. Three real roots, each with a different value B. One real root and to complex roots C. Three real roots, two of which are equal in value D. 2 real roots and one complex root?

Answer 1

Answer:

The correct option is C.

Step-by-step explanation:

The given cubic equation is

$x^3-5x^2+3x+9=0$

According to the rational root theorem 1 and -1 are possible rational roots of all polynomial.

At x=-1, the value of function 0. Therefore (x+1) is the factor of polynomial and -1 is a real root.

Use synthetic division to find the remaining polynomial.

$(x+1)(x^2-6x+9)=0$

$(x+1)(x^2-2(3)x+3^2)=0$

Using $(a-b)^2=a^2-2ab+b^2$

$(x+1)(x-3)^2)=0$

USe zero product property and equate each factor equal to 0.

$x=-1,x=3,x=3$

Therefore the equation have three real roots out of which the value of two roots are same.

Option C is correct.

Answer 2

The correct answer is letter C, because you have to calculate de discriminating which is 18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2 = 0
For your equation a=1, b=-5, c=3, d=9
So when the discriminating is equal zero, the equation has 3 real roots, two of which are equal in value.
You can prove this solving your equation using Ruffini's Rule and you will get the roots are: 3, 3, and -1