Question

Complete solution of inequalities (x-4)(3x-2)(4x+4)/(x+2)^2>0 and x^2-9 >_0

A) [3, infinity)
B) (4, infinity)
C) (1,2/3) U (4, infinity)
D) (-infinity, -3] U [3, infinity)​

Answer 1

Consider,

$\small\longrightarrow\dfrac{(x-4)(3x-2)(4x+4)}{(x+2)^2}>0$

Since denominator is non - zero,

$\small\Longrightarrow x\in\mathbb{R}-\{-2\}\quad\dots(i)$

Taking 4x + 4 = 4(x + 1),

$\small\longrightarrow\dfrac{4(x-4)(3x-2)(x+1)}{(x+2)^2}>0$

$\small\longrightarrow\dfrac{(x-4)(3x-2)(x+1)}{(x+2)^2}>0$

Since the denominator is always positive for every x ≠ -2,

$\small\longrightarrow(x-4)(3x-2)(x+1)>0$

For equality the roots are -1, 2/3 and 4 (3 real roots) in ascending order.

To satisfy the inequality there are two conditions.

• All the terms (x - 4), (3x - 2) and (x + 1) must be positive, i.e., no term must be negative.

• Exactly 2 terms among them must be negative.

For the first condition x must be greater than the largest root among them, i.e., 4.

$\small\Longrightarrow x\in(4,\ \infty)\quad\dots(ii)$

For the second condition x must be less than the second largest root but greater than the third largest root, i.e., x must lie exclusively between -1 and 2/3.

$\small\Longrightarrow x\in\left(-1,\ \dfrac{2}{3}\right)\quad\dots(iii)$

Taking (i) ∧ [(ii) ∨ (iii)] we get,

$\small\Longrightarrow x\in\left(-1,\ \dfrac{2}{3}\right)\cup(4,\ \infty)\quad\dots(1)$

[(i) is intersected to the union of (ii) and (iii) because (i) is common to (ii) and (iii) each, else (i) is not another particular case to (ii) and (iii) each.]

Consider,

$\small\longrightarrow x^2-9\geq0$

$\small\longrightarrow(x+3)(x-3)\geq0$

$\small\Longrightarrow x\in(-\infty,\ -3]\cup[3,\ \infty)\quad\dots(2)$

Taking (1) ∧ (2) we get,

$\small\text{ \longrightarrow\underline{\underline{x\in(4,\ \infty)}} }$

Hence (B) is the answer.