(A) Given,$a, b$ and $c$ are the roots of the equation $x^3+qx+r=0$. From Vieta's formulas: $a+b+c=0$ $ab+bc+ca=q$ $abc=-r$ Since $a+b+c=0$,we have $(

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

(A) Given,$a, b$ and $c$ are the roots of the equation $x^3+qx+r=0$. From Vieta's formulas: $a+b+c=0$ $ab+bc+ca=q$ $abc=-r$ Since $a+b+c=0$,we have $(a+b+c)^2=0$. $a^2+b^2+c^2+2(ab+bc+ca)=0$ $a^2+b^2+c^2+2(q)=0$ $a^2+b^2+c^2=-2q$ Now,consider the expression $(a-b)^2+(b-c)^2+(c-a)^2$: $= a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca$ $= 2(a^2+b^2+c^2) - 2(ab+bc+ca)$ $= 2(-2q) - 2(q)$ $= -4q - 2q$ $= -6q$

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Relation between roots and coefficients

Medium

If $a, b$ and $c$ are the roots of $x^3+qx+r=0$,then $(a-b)^2+(b-c)^2+(c-a)^2=$ (in $q$)

AP EAMCET 2018 All AP EAMCET

A
$-6$
B
$-4$
C
$6$
D
$4$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Mathematics Questions

Chapter

2 Complex Numbers and Quadratic Equations: Quadratic Equations

Subtopic

Relation between roots and coefficients

Previous Year

AP EAMCET 2018 Paper All AP EAMCET years →

If the roots of the equation $2x^2 - 3x + 5 = 0$ are the reciprocals of the roots of the equation $ax^2 + bx + 2 = 0$,then:

Easy

View Solution

If the roots of the equation $x^3-6x^2+11x-6=0$ are $\alpha, \beta$ and $\gamma$,then the equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is:

MediumAP EAMCET 2020

View Solution

If $\alpha, \beta, \gamma$ are the real roots of the equation $18x^3-15x^2-4x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$,then $\alpha+\beta^2+\gamma^3=$

EasyTS EAMCET 2023

View Solution

If a root of the given equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ is $1$,then the other root will be

Easy

View Solution

Let $f(n)=A(-2)^n+B(-3)^n$ for all $A, B \in \mathbb{R}$ and $n \in \mathbb{N}-\{1, 2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

EasyTS EAMCET 2020

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

If $a, b$ and $c$ are the roots of $x^3+qx+r=0$,then $(a-b)^2+(b-c)^2+(c-a)^2=$ (in $q$)

Similar Questions

If the roots of the equation $2x^2 - 3x + 5 = 0$ are the reciprocals of the roots of the equation $ax^2 + bx + 2 = 0$,then:

If the roots of the equation $x^3-6x^2+11x-6=0$ are $\alpha, \beta$ and $\gamma$,then the equation whose roots are $\alpha^2, \beta^2, \gamma^2$ is:

If $\alpha, \beta, \gamma$ are the real roots of the equation $18x^3-15x^2-4x+4=0$ such that $\alpha=\beta$ and $\alpha>\gamma$,then $\alpha+\beta^2+\gamma^3=$

If a root of the given equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ is $1$,then the other root will be

Let $f(n)=A(-2)^n+B(-3)^n$ for all $A, B \in \mathbb{R}$ and $n \in \mathbb{N}-\{1, 2\}$. If $f(n)+a f(n-1)+b f(n-2)=0$, then $(a+b)(b-a)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module