Consider the equation x^2 + alpha x + beta = | vedclass

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

Question

(A) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + \alpha x + \beta = 0$.By Vieta's formulas:Sum of roots: $\alpha + \beta = -\alpha

Vedclass · Accepted Answer

inequality is satisfied by exactly two integral values of $y$

Vedclass · Answer

(A) Given that $\alpha$ and $\beta$ are roots of the equation $x^2 + \alpha x + \beta = 0$.By Vieta's formulas:Sum of roots: $\alpha + \beta = -\alpha \implies 2\alpha + \beta = 0$  .......$(1)$Product of roots: $\alpha \beta = \beta \implies \beta(\alpha - 1) = 0$  .......$(2)$From $(2)$,either $\beta = 0$ or $\alpha = 1$.Case $I$: If $\beta = 0$,then from $(1)$,$2\alpha = 0 \implies \alpha = 0$. But the problem states $\alpha 
eq \beta$,so this is rejected.Case $II$: If $\alpha = 1$,then from $(1)$,$2(1) + \beta = 0 \implies \beta = -2$. Here $\alpha 
eq \beta$ holds.Now substitute $\alpha = 1$ and $\beta = -2$ into the inequality $| |y - (-2)| - 1 | $| |y + 2| - 1 | $-1 $0 This implies $|y + 2| $-2 $-4 Thus,$y \in (-4, -2) \cup (-2, 0)$.Checking option $D$: For $x^2 + x - 2 > 0$,the roots are $x = -2, 1$. The parabola opens upward,so $x^2 + x - 2 > 0$ for $x \in (-\infty, -2) \cup (1, \infty)$. This does not hold for $x \in [-1, 0]$.

Consider the equation $x^2 + \alpha x + \beta = 0$ having roots $\alpha, \beta$ such that $\alpha \neq \beta$. Also consider the inequality $| |y - \beta| - \alpha | < \alpha$,then:

Similar Questions

Construct a quadratic equation whose roots have the sum $= 6$ and product $= -16$.

The least positive value of $a$ for which the equation $2x^{2} + (a - 10)x + \frac{33}{2} = 2a$ has real roots is

If $a+b=1,$ find the value of $a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}$.

If $\alpha, \beta$ are the roots of the equation $a x^{2}+b x+b=0,$ then $\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{b}{a}}=$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module