Number of integers n for which 3x^3 −25x+n=0 has three real roots is ?


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Question Number 46974 by rahul 19 last updated on 03/Nov/18

$N u m b e r o f i n t e g e r s n f o r w h i c h$ $3 x^{3} - 25 x + n = 0 h a s t h r e e r e a l r o o t s i s ?$
Commented by rahul 19 last updated on 03/Nov/18

$A n s . i s 55 b u t i^{'} m g e t t i n g i n f i n i t e . . .$ $p l s s c h e c k m y m e t h o d . . . .$ $f (x) = 3 x^{3} - 25 x + n = 0$ $f^{'} (x) = 9 x^{2} - 25 = 0$ $\Rightarrow x = \mp \sqrt{\frac{5}{3}} \Rightarrow f^{'} (x) h a s 2 r e a l r o o t s$ $\Rightarrow d e f i n i t e l y f (x) w i l l 3 r e a l r o o t s . . .$ $h e n c e n c a n b e a n y i n t e g e r \to i n f i n i t e . . .$
	Answered by MJS last updated on 03/Nov/18

	$x^{3} + p x + q = 0$ $D = \frac{p^{3}}{27} + \frac{q^{2}}{4}$ $D < 0 \Leftrightarrow 3 real roots$ $in this case$ $x^{3} - \frac{25}{3} x + \frac{n}{3} = 0$ $D = \frac{n^{2}}{36} - \frac{15625}{729} < 0 \Rightarrow - \frac{250}{9} ⩽ n ⩽ \frac{250}{9}$ $n \in Z \Rightarrow - 27 ⩽ n ⩽ 27 \Rightarrow 55 integer values for n$
	Commented by rahul 19 last updated on 04/Nov/18
	thanks sir ��
	Answered by ajfour last updated on 03/Nov/18
	$f(x)=3x^3 −25x f ′(x)=9x^2 −25 f ′(x)=0 ⇒ x= ±(5/3), 0 f((5/3))=3((5/3))^3 −25((5/3)) = ((125)/9)−((125)/3) = − ((250)/9) so having three real roots we can raise or lower f(x) by n< ((250)/9) ⇒ n∈ (−((250)/9) , ((250)/9)) ⇒ n∈ {−27,−26,....,−1,0, 1,..27} ⇒ 55 integral values.$
	$f (x) = 3 x^{3} - 25 x$ $f^{'} (x) = 9 x^{2} - 25$ $f^{'} (x) = 0 \Rightarrow x = \pm \frac{5}{3}, 0$ $f (\frac{5}{3}) = 3 {(\frac{5}{3})}^{3} - 25 (\frac{5}{3})$ $= \frac{125}{9} - \frac{125}{3} = - \frac{250}{9}$ $s o h a v i n g t h r e e r e a l r o o t s w e$ $c a n r a i s e o r l o w e r f (x) b y$ $n < \frac{250}{9}$ $\Rightarrow n \in (- \frac{250}{9}, \frac{250}{9})$ $\Rightarrow n \in {- 27, - 26, . . . ., - 1, 0, 1, . . 27}$ $\Rightarrow 55 i n t e g r a l v a l u e s .$
	Commented by rahul 19 last updated on 04/Nov/18
	thanks sir ��
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