The sum of all distinct integral values of $\alpha$ such that the equation $x^2 - \alpha x + \alpha + 1 = 0$ has integral roots is equal to:

  • A
    $2$
  • B
    $4$
  • C
    $3$
  • D
    None of these

Explore More

Similar Questions

Let $p, q, r \in \mathbb{R}$ and $r > p > 0$. If the quadratic equation $px^2 + qx + r = 0$ has two complex roots $\alpha$ and $\beta$,then what is the value of $|\alpha| + |\beta|$?

Difficult
View Solution

For how many values of $k$ is the equation $(1 + 2k)x^2 + (1 - 2k)x + (1 - 2k) = 0$ a perfect square?

Which of the following six statements are true about the cubic polynomial $P(x) = 2x^3 + x^2 + 3x - 2$?
$(i)$ It has exactly one positive real root.
$(ii)$ It has either one or three negative roots.
$(iii)$ It has a root between $0$ and $1$.
$(iv)$ It must have exactly two real roots.
$(v)$ It has a negative root between $-2$ and $-1$.
$(vi)$ It has no complex roots.

If $a$ is a positive integer such that the roots of the equation $7x^2 - 13x + a = 0$ are rational numbers,then the smallest possible value of $a$ is

The value of $3+\frac{1}{4+\frac{1}{3+\frac{1}{4+\frac{1}{3+\ldots \infty}}}}$ is equal to

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