Number-of-integers-n-for-which-3x-3-25x-n-0-has-three-real-roots-is-

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 \dots

p l s s c h e c k m y m e t h o d \dots .

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 \dots

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 \dots

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) = 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, \dots ., - 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 ��