Question

(x2 + 3x + 1) (x2 + 3x - 3) > 5​

Answer 1

Take y = x^2 + 3x

Now, (y+1)(y-3)>=5

y^2 -2y -8 >=0

(y+2)(y-4)>=0

Now y = (-inf,-2) (4,inf)

x^2 + 3x <= -2

(x+2)(x+1)<=0

x belongs from (-1,-2)

x^2 + 3x >=4

(x+4)(x-1)>=0

Hence, x= (-inf,-4) (1,inf)

These are the solutions

THanks

Answer 2

Answer:

x ∈​ (-∞ , -4] U [-2 , -1] U [1 , ∞)

Step-by-step explanation:

Let   x² + 3x = y

Now, (y + 1)(y - 3) ≥ 5

y² +y -3y -3 ≥ 5

y²-2y -8 ≥ 0

(y+2)(y-4)≥0

Now y = -2 or 4

at y= -2

x² + 3x ≥ -2

( x+2 )( x+1 ) ≥ 0

x belongs from (-1,-2)

at y= 4

x² + 3x ≥ -4

( x+4 )( x-1 ) ≥ 0

x belongs from (-4,1)

Put no. -4 , -2 , -1 , 1 on number line

 

____+__________-___________+___________-__________+___

                  -4                    -2                       -1                        1                

∴ (x2 + 3x + 1) (x2 + 3x - 3) ≥ 5

x ∈​ (-∞ , -4] U [-2 , -1] U [1 , ∞)

Thanks