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Question Number 47019 by ajfour last updated on 04/Nov/18

Commented by ajfour last updated on 04/Nov/18

$I f a t i t s r o o t s x = - 3 a n d x = 2$ $t h e b i q u a d r a t i c a l s o h a s i t s$ $p o i n t s o f i n f l e x i o n, f i n d i t s$ $o t h e r t w o r o o t s α a n d β .$
Commented by ajfour last updated on 04/Nov/18

$s h o w n g r a p h i s j u s t t o h e l p$ $r o u g h l y d e p i c t t h e q u e s t i o n!$
	Answered by MJS last updated on 04/Nov/18

	$y = {a x}^{4} + {b x}^{3} + {c x}^{2} + d x + e$ $y^{″} = 12 {a x}^{2} + 6 b x + 2 c$ $(1) 81 a - 27 b + 9 c - 3 d + e = 0$ $(2) 16 a + 8 b + 4 c + 2 d + e = 0$ $(3) 108 a - 18 b + 2 c = 0$ $(4) 48 a + 12 b + 2 c = 0$ $this leads to$ $y = {a x}^{4} + 2 {a x}^{3} - 36 {a x}^{2} - 37 a x + 186 a$ $a (x + 3) (x - 2) (x^{2} + x - 31) = 0$ $\Rightarrow α = - \frac{1}{2} - \frac{5 \sqrt{5}}{2}; β = - \frac{1}{2} + \frac{5 \sqrt{5}}{2}$
	Commented by MJS last updated on 04/Nov/18

	$(x + 3) (x - 2) (x + \frac{1}{2} + \frac{5 \sqrt{3}}{2}) (x + \frac{1}{2} - \frac{5 \sqrt{3}}{2}) =$ $= x^{4} + 2 x^{3} - \frac{47}{2} x^{2} - \frac{49}{2} x + 111$ $\frac{d^{2}}{{d x}^{2}} [x^{4} + 2 x^{3} - \frac{47}{2} x^{2} - \frac{49}{2} x + 111] = 12 x^{2} + 12 x - 47$ $x = - 3 \Rightarrow 12 x^{2} + 12 x - 47 = 25$ $x = 2 \Rightarrow 12 x^{2} + 12 x - 47 = 25$ $so something went wrong$
	Commented by ajfour last updated on 04/Nov/18

	$y e s S i r, i g o t m y e r r o r, t h a n k s!$
	Commented by MJS last updated on 04/Nov/18

	$as always, {you}^{'} re welcome$
	Commented by peter frank last updated on 04/Nov/18

	$pls help QN 46813$
	Answered by ajfour last updated on 04/Nov/18

	$y = a (x^{2} + x - 6) (x^{2} + p x + q)$ $y^{'} = a (2 x + 1) (x^{2} + p x + q)$ $+ a (x^{2} + x - 6) (2 x + p)$ $y^{″} = a [2 (x^{2} + p x + q) + (2 x + 1) (2 x + p)$ $+ (2 x + 1) (2 x + p) + 2 (x^{2} + x - 6)]$ $\Rightarrow y^{″} = 2 a [x^{2} + p x + q$ $+ (2 x + 1) (2 x + p)$ $+ x^{2} + x - 6]$ $\Rightarrow y^{″} = a [6 x^{2} + 3 (p + 1) x + (q + p - 6)]$ $a s i t s r o o t s a r e x = - 3, 2$ $\Rightarrow - \frac{(p + 1)}{2} = - 1 \Rightarrow p = 1$ $a n d q + p - 6 = - 36$ $\Rightarrow q = - 31$ $h e n c e r o o t s o f x^{2} + p x + q = 0$ $\Rightarrow x^{2} + x - 31 = 0$ $t h a t i s α, β a r e$ $= \frac{- 1 \pm \sqrt{1 + 124}}{2} = \frac{- 1 \pm 5 \sqrt{5}}{2} .$
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