Question

solve:
$(x {2} + 3x + 1)(x {2} + 3x - 3) \geqslant 5$
find the values of x​

Answer 1

$\large\underline{\sf{Solution-}}$

Given inequality is

$\rm :\longmapsto\:( {x}^{2} + 3x + 1)( {x}^{2} + 3x - 3) \geqslant 5$

Let we assume that,

$\red{\rm :\longmapsto\: {x}^{2} + 3x = y}$

So, above inequality can be rewritten as

$\rm :\longmapsto\:(y + 1)(y - 3) \geqslant 5$

$\rm :\longmapsto\: {y}^{2} - 3y + y - 3 \geqslant 5$

$\rm :\longmapsto\: {y}^{2} - 2y - 8 \geqslant 0$

$\rm :\longmapsto\: {y}^{2} - 4y + 2y - 8 \geqslant 0$

$\rm :\longmapsto\:y(y - 4) + 2(y - 4) \geqslant 0$

$\rm :\longmapsto\:(y - 4)(y + 2) \geqslant 0$

On substituting back the value of y, we get

$\rm :\longmapsto\:( {x}^{2} + 3x - 4)( {x}^{2} + 3x + 2) \geqslant 0$

$\rm :\longmapsto\:\bigg( {x}^{2} + 4x - x - 4 \bigg) \bigg( {x}^{2} + 2x + x + 2 \bigg) \geqslant 0$

$\rm :\longmapsto\:\bigg(x(x + 4) - 1(x + 4) \bigg)\bigg(x(x + 2) + 1(x + 2) \bigg) \geqslant 0$

$\rm :\longmapsto\:(x + 4)(x - 1)(x + 2)(x + 1) \geqslant 0$

So, breaking points are - 4, - 2, - 1, 1

Let check the interval for the sign.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x \leqslant - 4 & \sf + \\ \\ \sf - 4 \leqslant x \leqslant - 2 & \sf - \\ \\ \sf - 2 \leqslant x \leqslant - 1 & \sf + \\ \\ \sf - 1 \leqslant x \leqslant 1 & \sf - \\ \\ \sf x \geqslant 1 & \sf + \end{array}} \\ \end{gathered}$

So, we concluded that

$\bf\implies \:x \in \: ( - \infty , - 4] \cup \: [ - 2, - 1] \cup \: [ 1, \infty )$

Answer 2

✫SOLUTION✫

(x² + 3x +1)(x² +3x -3) ≥ 5

let x² + 3x = P

(P +1)( P-3) ≥ 5

P² -2P-3 ≥ 5

P² -2P-3 -5 20

P² -2P-8 ≥ 0

P² - 4P +2P -8 20

P(P-4) +2( P-4) ≥ 0

(P +2)( P-4) 20

P24 and P≤ -2

now,

x² +3x ≥ 4 and x² +3x≤-2

x² +3x -4 ≥ 0

x² +4x -x -4 20

(x +4)(x-1) 20

x ≥1 and x≤ -4

Again,

x² +3x≤-2

x² + 3x +2 ≤0

(x + 1)(x+2) ≤ 0

-2 ≤x≤-1

Hence, solutions is x € [-2, -1] U [1, ∞ ) U (-∞, 4 ]

THANKS!!