(D) Let $y = \frac{(x+2)(x+5)}{x+6}$.$xy + 6y = x^2 + 7x + 10$$x^2 + (7-y)x + (10-6y) = 0$For $x$ to be real, the discriminant $\Delta \geq 0$.$(7-y)^

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(D) Let $y = \frac{(x+2)(x+5)}{x+6}$.$xy + 6y = x^2 + 7x + 10$$x^2 + (7-y)x + (10-6y) = 0$For $x$ to be real, the discriminant $\Delta \geq 0$.$(7-y)^2 - 4(10-6y) \geq 0$$y^2 - 14y + 49 - 40 + 24y \geq 0$$y^2 + 10y + 9 \geq 0$$(y+1)(y+9) \geq 0$Thus, $y \in (-\infty, -9] \cup [-1, \infty)$.The values of $y$ that do not lie in the range are in the interval $(-9, -1)$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic inequations and Newton Formula

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For $x \in R-\{-6\}$, the value of $\frac{(x+2)(x+5)}{(x+6)}$ does not lie in the interval

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A
$[-9,-1]$
B
$[-5,-2]$
C
$(-5,-2)$
D
$(-9,-1)$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic inequations and Newton Formula

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For $x \in R-\{-6\}$, the value of $\frac{(x+2)(x+5)}{(x+6)}$ does not lie in the interval

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