Question

Solve (x² + 3x + 1)(x² + 3x - 3) ≥ 5
Please solve it!!

Answer 1

(x² + 3x +1)(x² +3x -3) ≥ 5

let x² + 3x = P

(P +1)( P -3) ≥ 5

P² -2P -3 ≥ 5

P² -2P -3 -5 ≥ 0

P² -2P -8 ≥ 0

P² - 4P +2P -8 ≥ 0

P( P -4) +2( P -4) ≥ 0

(P +2)( P -4) ≥ 0

P≥ 4 and P≤ -2

now,

x² +3x ≥ 4 and x² +3x ≤ -2

x² +3x -4 ≥ 0

x² +4x -x -4 ≥0

(x +4)(x -1) ≥0

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 ]

Answer 2

I kind of learnt this inequality concept so I want to try this out.

(x² + 3x +1)(x² +3x -3) ≥ 5

Here x² + 3x is common so
Let x² + 3x = Y

(Y +1)( y -3) ≥ 5
Y² -3Y+Y -3 ≥ 5
Y² - 2Y - 3 ≥ 5
Y² -2Y -3 -5 ≥ 0
Y² -2Y -8 ≥ 0
Y² - 4Y +2Y -8 ≥ 0
Y( Y -4) +2( Y -4) ≥ 0
(Y +2)( Y - 4) ≥ 0

So,

Y ≥ 4 and Y ≤  -2

x² +3x ≥ 4 and x² +3x ≤ -2
x² +3x -4 ≥ 0
x² +4x -x -4 ≥0
(x +4)(x -1) ≥0
x ≥ 1 and x ≤ -4
Now,
x² +3x ≤ -2
x² +3x +2 ≤ 0
(x +1)(x +2) ≤ 0
-2 ≤ x ≤ -1

Required solution ---
x € [ -2, -1 ] U [ 1, ∞ ) U ( -∞ , 4 ]

Hope This Helps You