solve-x-2-3x-1-x-2-x-1-lt-3-

Question Number 7598 by Rohit last updated on 05/Sep/16

s o l v e ∣ \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} ∣< 3

Answered by Rasheed Soomro last updated on 06/Sep/16

∣ \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} ∣< 3

\pm \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} < 3

\frac{x^{2} - 3 x - 1}{x^{2} + x + 1} < 3 ∣ - \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} < 3

\frac{x^{2} - 3 x - 1}{x^{2} + x + 1} < 3 ∣ \frac{x^{2} - 3 x - 1}{x^{2} + x + 1} > - 3

x^{2} + x + 1 > 0 [S e e t h e r e a s o n i n t h e c o m m e n t b y S a n d y b e l o w

o r s e e t h e e x p l a n a t i o n f o r t h i s i n t h e a n s w e r b y Y o z z i a .]

x^{2} - 3 x - 1 < 3 x^{2} + 3 x + 3 ∣ x^{2} - 3 x - 1 > - 3 x^{2} - 3 x - 3

2 x^{2} + 6 x + 4 > 0 ∣ 4 x^{2} + 2 > 0

x^{2} + 3 x + 2 > 0 ∣ 2 x^{2} + 1 > 0

(x + 2) (x + 1) > 0 ∣ x^{2} > - \frac{1}{2} \Rightarrow x^{2} > 0 \Rightarrow x > 0 \lor x < 0

(x + 2 > 0 \land x + 1 > 0) o r (x + 2 < 0 \land x + 1 < 0)

x > - 2 \land x > - 1 o r x < - 2 \land x < - 1

x > - 1 o r x < - 2

Commented by sandy_suhendra last updated on 05/Sep/16

(x^{2} + x + 1) i s a l w a y s p o s i t i v e,

b e c a u s e a > 0 a n d D = b^{2} - 4 a c = 1^{2} - 4.1.1 = - 3 \Rightarrow D < 0

Commented by Rasheed Soomro last updated on 05/Sep/16

{TH}_{n}^{α} KS!