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Polynomial and Its General Form

Solve: |x2+...

Question

Solve: $| (x^{2} + 2 x + 2) + (3 x + 7) | < | x^{2} + 2 x + 2 | + | 3 x + 7 |$ .

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Solution

$∣ ∣ x^{2} + 2 x + 2 + 3 x + 7 ∣ ∣ < ∣ ∣ x^{2} + 2 x + 2 ∣ ∣ + | 3 x + 7 |$
$∣ ∣ x^{2} + 5 x + 9 ∣ ∣ < ∣ ∣ x^{2} + 2 x + 2 ∣ ∣ + | 3 x + 7 |$
now, $(x^{2} + 5 x + 9)$ and $(x^{2} + 5 x + 9)$ both have D<0 and hence are >0
$x^{2} + 5 x + 9 < x^{2} + 2 x + 2$
$| 3 x + 7 | > 3 x + 7$
(i) For $x < \frac{- 7}{3}$ , $3 x + 14$
$6 x + 14 < 0$
$x < \frac{- 7}{3}$
$x ε (- \infty, \frac{- 7}{3})$
(ii) For $x > \frac{- 7}{3}$
$3 x + 7 > 3 x + 7$
$x ε (- \infty, \frac{- 7}{3})$

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