(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation. Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$. We know that $(\alpha+\beta)^2

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

(D) Let $\alpha$ and $\beta$ be the roots of the quadratic equation. Given $\alpha+\beta=11$ and $\alpha^2+\beta^2=61$. We know that $(\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta$. Substituting the values: $(11)^2 = 61 + 2\alpha\beta$. $121 = 61 + 2\alpha\beta$ $\Rightarrow 2\alpha\beta = 60$ $\Rightarrow \alpha\beta = 30$. The quadratic equation is given by $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$. Substituting the values: $x^2 - 11x + 30 = 0$.

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

Easy

The quadratic equation whose sum of the roots is $11$ and sum of squares of the roots is $61$ is

AP EAMCET 2022 All AP EAMCET

A
$x^2+11x-30=0$
B
$x^2+11x+30=0$
C
$x^2-11x-30=0$
D
$x^2-11x+30=0$

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 2022 Paper All AP EAMCET years →

If $\alpha, \beta, \gamma$ are the roots of $2x^3 - 2x - 1 = 0$,then $(\Sigma \alpha \beta)^2$ is equal to

MediumTS EAMCET 2002

View Solution

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 4 = 0$,then $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$ is equal to

Easy

View Solution

The value of $a$ for which one root of the quadratic equation $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$ is twice as large as the other,is

MediumAIEEE 2003

View Solution

If $\alpha, \beta$ are the roots of $x^2+p x+q=0$,then the values of $\alpha^3+\beta^3$ and $\alpha^4+\alpha^2 \beta^2+\beta^4$ are respectively ...... and ......

MediumAP EAMCET 2020

View Solution

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 3x + 5 = 0$,then find the quadratic equation whose roots are $(\alpha^2 - 3\alpha + 7)$ and $(\beta^2 - 3\beta + 7)$.

Difficult

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

The quadratic equation whose sum of the roots is $11$ and sum of squares of the roots is $61$ is

Similar Questions

If $\alpha, \beta, \gamma$ are the roots of $2x^3 - 2x - 1 = 0$,then $(\Sigma \alpha \beta)^2$ is equal to

If $\alpha$ and $\beta$ are the roots of the equation $x^2 + 2x + 4 = 0$,then $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$ is equal to

The value of $a$ for which one root of the quadratic equation $(a^2 - 5a + 3)x^2 + (3a - 1)x + 2 = 0$ is twice as large as the other,is

If $\alpha, \beta$ are the roots of $x^2+p x+q=0$,then the values of $\alpha^3+\beta^3$ and $\alpha^4+\alpha^2 \beta^2+\beta^4$ are respectively ...... and ......

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 - 3x + 5 = 0$,then find the quadratic equation whose roots are $(\alpha^2 - 3\alpha + 7)$ and $(\beta^2 - 3\beta + 7)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module