Solve 1st: 3x^2 −2x−11> 0 2nd: x^4 + 4x^3 − 12x^2 ≤ 0


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Question Number 178553 by Acem last updated on 18/Oct/22

$S o l v e 1 s t : 3 x^{2} - 2 x - 11 > 0$ $2 n d : x^{4} + 4 x^{3} - 12 x^{2} ⩽ 0$
	Answered by a.lgnaoui last updated on 18/Oct/22

	$1 s t e x t r e m u m s := (\frac{1 - \sqrt{34}}{3}; \frac{1 + \sqrt{34}}{3}) \Rightarrow 3 x^{2} - 2 x - 11 > 0 \Leftrightarrow x \in] - \infty, \frac{1 - \sqrt{34}}{3} [\cup] \frac{1 + \sqrt{34}}{3}, + \infty$ $2 n d e x t r e m u m s : (- (1 + \sqrt{7}), 0, + (1 + \sqrt{7}) \Rightarrow 2 x^{4} + 4 x^{3} - 12 x^{2} ⩽ 0$ $\Leftrightarrow x \in [- (1 + \sqrt{7}), - 1 + \sqrt{7})]$
	Commented byAcem last updated on 18/Oct/22

	$G o o d S i r! t h e 1 s t i s r i g h t, t h e 2 n d i s x \in [- 6, 4]$ $t r y i t a g a i n, m a y b e i^{'} m w r o n g$ $A m s o r r y i n o t e d t h e f i r s t t e r m a s w r o n g$ ${i t}^{'} s x^{4} n o t 2 x^{4}$
	Answered by mr W last updated on 18/Oct/22

	$(2)$ $2 x^{2} (x^{2} + 2 x - 6) ⩽ 0$ $\Rightarrow x^{2} + 2 x - 6 ⩽ 0$ $\Rightarrow - 1 - \sqrt{7} ⩽ x ⩽ - 1 + \sqrt{7}$
	Commented byAcem last updated on 18/Oct/22

	$A m s o r r y i n o t e d t h e f i r s t t e r m a s w r o n g$ ${i t}^{'} s x^{4} n o t 2 x^{4}$
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