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Question Number 114461 by Jamshidbek2311 last updated on 19/Sep/20

Commented by bemath last updated on 19/Sep/20

$∣ (x + 2) (x + 5) ∣ + 10 ∣ (x - 2) (x + 2) >∣ (2 x + 3) (x + 2) ∣$ $∣ x + 2 ∣ {∣ x + 5 ∣ - 10 ∣ x - 2 ∣ - ∣ 2 x + 3 ∣} > 0$ $n o w i t s c a n s o l v e$
	Answered by 1549442205PVT last updated on 19/Sep/20
	$Solve:∣x^2 +7x+10∣+∣x^2 −4∣>∣2x^2 +7x+6∣(1) x^2 +7x+10=(x+2)(x+5)≥0⇔x∈(−∞;−5)∪(−2;+∞) (x^2 −4)≥0⇔x∈(−∞;−2)∪(2;+∞) 2x^2 +7x+6≥0⇔(2x+3)(x+2)≥0⇔x∈(−∞;−2)∪(−(3/2);+∞) Hence we have followingtablet: determinant ((x,,(−5),,(−2),,(−(3/2)),,2,),(A,(x^2 +7x+10),0,(−x^2 −7x−10),0,(x^2 +7x+10),,(x^2 +7x+10),,(x^2 +7x+10)),(B,(x^2 −4),,(x^2 −4),,(4−x^2 ),,(4−x^2 ),0,(x^2 −4)),(C,(2x^2 +7x+6),,(2x^2 +7x+6),0,(−2x^2 −7x−6),0,(2x^2 +7x+6),,(2x^2 +7x+6))) From above we get i)For x∈(−∞;−5] (1)⇔2x^2 +7x+7>2x^2 +7x+6⇔has no roots ii)For x∈(−5;−2] (1)⇔−7x−14>2x^2 +7x+6 ⇔2x^2 +14x+20<0⇔x^2 +7x+10<0 ⇔(x+2)(x+5)<0⇔x∈ (−5;−2) we has the roots:x∈(−5;−2) iii)For x∈(−2;−(3/2)] (1)⇔7x+14>−2x^2 −7x−6 ⇔2x^2 +14x+20>0⇔x^2 +7x+10>0 ⇔(x+2)(x+5)>0⇔x∈(−∞;−5)∪(−2;+∞) we have the roots x∈(−2;−(3/2)) iv)For x∈(((−3)/2);2] (1)⇔7x+14>2x^2 +7x+6 ⇔2x^2 −8<0⇔x^2 −4<0⇔x∈(−2;2) we have the roots x∈(((−3)/2);2) v)For x∈(2;+∞) (1)⇔2x^2 +7x+6>2x^2 +7x+6 This inequality has no roots Combining five cases we get the roots of given inequality are: x∈(−5;−2)∪(−2;−(3/2))∪(−(3/2);2) or∈(−5;2)\{−2;−(3/2)}$
	$Solve :∣ x^{2} + 7 x + 10 ∣ + ∣ x^{2} - 4 ∣>∣ {2 x}^{2} + 7 x + 6 ∣ (1)$ $x^{2} + 7 x + 10 = (x + 2) (x + 5) ⩾ 0 \Leftrightarrow x \in (- \infty; - 5) \cup (- 2; + \infty)$ $(x^{2} - 4) ⩾ 0 \Leftrightarrow x \in (- \infty; - 2) \cup (2; + \infty)$ ${2 x}^{2} + 7 x + 6 ⩾ 0 \Leftrightarrow (2 x + 3) (x + 2) ⩾ 0 \Leftrightarrow x \in (- \infty; - 2) \cup (- \frac{3}{2}; + \infty)$ $Hence we have followingtablet :$ $\| \begin{matrix} x & - 5 & - 2 & - \frac{3}{2} & 2 \\ A & x^{2} + 7 x + 10 & 0 & - x^{2} - 7 x - 10 & 0 & x^{2} + 7 x + 10 & x^{2} + 7 x + 10 & x^{2} + 7 x + 10 \\ B & x^{2} - 4 & x^{2} - 4 & 4 - x^{2} & 4 - x^{2} & 0 & x^{2} - 4 \\ C & {2 x}^{2} + 7 x + 6 & {2 x}^{2} + 7 x + 6 & 0 & - {2 x}^{2} - 7 x - 6 & 0 & {2 x}^{2} + 7 x + 6 & {2 x}^{2} + 7 x + 6 \end{matrix} \|$ $From above we get$ $i) For x \in (- \infty; - 5]$ $(1) \Leftrightarrow {2 x}^{2} + 7 x + 7 > {2 x}^{2} + 7 x + 6 \Leftrightarrow has no roots$ $ii) For x \in (- 5; - 2]$ $(1) \Leftrightarrow - 7 x - 14 > {2 x}^{2} + 7 x + 6$ $\Leftrightarrow {2 x}^{2} + 14 x + 20 < 0 \Leftrightarrow x^{2} + 7 x + 10 < 0$ $\Leftrightarrow (x + 2) (x + 5) < 0 \Leftrightarrow x \in (- 5; - 2)$ $we has the roots : x \in (- 5; - 2)$ $iii) For x \in (- 2; - \frac{3}{2}]$ $(1) \Leftrightarrow 7 x + 14 > - {2 x}^{2} - 7 x - 6$ $\Leftrightarrow {2 x}^{2} + 14 x + 20 > 0 \Leftrightarrow x^{2} + 7 x + 10 > 0$ $\Leftrightarrow (x + 2) (x + 5) > 0 \Leftrightarrow x \in (- \infty; - 5) \cup (- 2; + \infty)$ $we have the roots$ $x \in (- 2; - \frac{3}{2})$ $iv) For x \in (\frac{- 3}{2}; 2]$ $(1) \Leftrightarrow 7 x + 14 > {2 x}^{2} + 7 x + 6$ $\Leftrightarrow {2 x}^{2} - 8 < 0 \Leftrightarrow x^{2} - 4 < 0 \Leftrightarrow x \in (- 2; 2)$ $we have the roots x \in (\frac{- 3}{2}; 2)$ $v) For x \in (2; + \infty)$ $(1) \Leftrightarrow {2 x}^{2} + 7 x + 6 > {2 x}^{2} + 7 x + 6$ $This inequality has no roots$ $Combining five cases we get the roots$ $of given inequality are :$ $x \in (- 5; - 2) \cup (- 2; - \frac{3}{2}) \cup (- \frac{3}{2}; 2)$ $or \in (- 5; 2) ∖ {- 2; - \frac{3}{2}}$
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