Class 11Mathematics2 Complex Numbers and Quadratic Equations: Quadratic EquationsSolution of quadratic equations and Nature of roots
EasyIf two roots of the equation $x^3 - 3x + 2 = 0$ are the same,then the roots are:
- A$2, 2, 3$
- B$1, 1, -2$
- C$-2, 3, 3$
- D$-2, -2, 1$
Explore More
Similar Questions
The product of the real roots of the equation $x^2 - |x| - 6 = 0$ is .......
Difficult
View SolutionLet $\alpha$ and $\beta$ be the roots of the quadratic equation $a x^2+b x+c=0$. Observe the lists given below:
List-$I$ List-$II$ $(i)$ $\alpha = \beta$ $(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ $(ii)$ $\alpha = 2\beta$ $(B)$ $2b^2 = 9ac$ $(iii)$ $\alpha = 3\beta$ $(C)$ $b^2 = 6ac$ $(iv)$ $\alpha = \beta^2$ $(D)$ $3b^2 = 16ac$ $(E)$ $b^2 = 4ac$ $(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$
The correct match of List-$I$ from List-$II$ is:
| List-$I$ | List-$II$ |
| $(i)$ $\alpha = \beta$ | $(A)$ $(ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$ |
| $(ii)$ $\alpha = 2\beta$ | $(B)$ $2b^2 = 9ac$ |
| $(iii)$ $\alpha = 3\beta$ | $(C)$ $b^2 = 6ac$ |
| $(iv)$ $\alpha = \beta^2$ | $(D)$ $3b^2 = 16ac$ |
| $(E)$ $b^2 = 4ac$ | |
| $(F)$ $(ac^2)^{1/3} + (a^2c)^{1/3} = b$ |
The correct match of List-$I$ from List-$II$ is:
$\alpha$ is a root of the equation $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$. If $-\frac{1}{2} < \alpha < 0$,then $\alpha = $
For what value of $k$ will the equation $x^2 - (3k - 1)x + 2k^2 + 2k = 0$ have equal roots?
Easy
View SolutionLet $f(x)$ be a quadratic expression such that $f(0)+f(1)=0$. If $f(-2)=0$,then
DifficultTS EAMCET 2017
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo