The sum of all non-integer roots of the equat | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$. By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Given the equation: $x^5-6x^4+11x^3-5x^2-3x+2=0$. By testing integer roots using the Rational Root Theorem,we find that $x=1$ and $x=2$ are roots. Dividing the polynomial by $(x-1)(x-2) = x^2-3x+2$,we get: $(x-1)(x-2)(x^3-3x^2+1)=0$. The non-integer roots are the roots of the cubic equation $x^3-3x^2+1=0$. Let the roots of this cubic equation be $\alpha, \beta, \gamma$. By Vieta's formulas,the sum of the roots is given by $-\frac{b}{a} = -\frac{-3}{1} = 3$. Thus,the sum of all non-integer roots is $3$.

The sum of all non-integer roots of the equation $x^5-6x^4+11x^3-5x^2-3x+2=0$ is

Similar Questions

Let $S = \{ \sin^2 2\theta : (\sin^4 \theta + \cos^4 \theta)x^2 + (\sin 2\theta)x + (\sin^6 \theta + \cos^6 \theta) = 0 \text{ has real roots} \}$. If $\alpha$ and $\beta$ are the smallest and largest elements of the set $S$,respectively,then $3((\alpha - 2)^2 + (\beta - 1)^2)$ equals:

If $-1$ is a twice repeated root of the equation $a(x^3+x^2)+bx+c=0$,then $a:b:c=$

The number of real solutions of the equation $(\frac{3}{2})^x = -x^2 + 5x - 10$ is:

Let $a, b, c \in \mathbb{R}$ and $\alpha, \beta$ be the real roots of the equation $ax^2 + bx + c = 0$. If $a < 0, b > 0, c > 0$ and $\alpha < \beta$,then:

If $a, b, c, d$ are positive real numbers such that $a + b + c + d = 2$,then $M = (a + b)(c + d)$ satisfies the relation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module