The least integral value alpha of x such that | vedclass

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

Question

(D) Given the inequality: $\frac{x - 5}{x^2 + 5x - 14} > 0$.Factor the denominator: $x^2 + 5x - 14 = (x + 7)(x - 2)$.So,the inequality is $\frac{x - 5

Vedclass · Accepted Answer

$\alpha^2 + 5\alpha - 6 = 0$

Vedclass · Answer

(D) Given the inequality: $\frac{x - 5}{x^2 + 5x - 14} > 0$.Factor the denominator: $x^2 + 5x - 14 = (x + 7)(x - 2)$.So,the inequality is $\frac{x - 5}{(x + 7)(x - 2)} > 0$.Using the wavy curve method (sign scheme),the critical points are $x = -7, 2, 5$.The expression is positive in the intervals $(-7, 2) \cup (5, \infty)$.The integers in the interval $(-7, 2)$ are $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.The integers in the interval $(5, \infty)$ are $\{6, 7, 8, \dots\}$.The least integral value $\alpha$ is $-6$.Checking the options for $\alpha = -6$:$(A)$ $(-6)^2 + 3(-6) - 4 = 36 - 18 - 4 = 14 
eq 0$.$(B)$ $(-6)^2 - 5(-6) + 4 = 36 + 30 + 4 = 70 
eq 0$.$(C)$ $(-6)^2 - 7(-6) + 6 = 36 + 42 + 6 = 84 
eq 0$.$(D)$ $(-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0$.Thus,$\alpha = -6$ satisfies option $(D)$.

The least integral value $\alpha$ of $x$ such that $\frac{x - 5}{x^2 + 5x - 14} > 0$ satisfies:

Similar Questions

If the product of the roots of the equation $(a + 1)x^2 + (2a + 3)x + (3a + 4) = 0$ is $2$,then the sum of the roots is:

If $\alpha$ and $\beta$ are the roots of $9x^2 + 6x + 1 = 0$,then the equation with the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + x - 1 = 0$
$II.$ $2y^{2} - 3y + 1 = 0$

Solve the given two equations and select the correct answer from the given options.
$I.$ $17 x^{2} + 48 x = 9$
$II.$ $13 y^{2} = 32 y - 21$

Solve the given two equations and select the correct option.
$I.$ $7x - 3y = 13$
$II.$ $5x + 4y = 40$

Vedclass Test Series

Exam Paper Generator

Online Exam Module