The-number-of-real-solutions-of-the-equation-4x-99-5x-98-4x-97-5x-96-4x-5-0-is-

Question Number 21321 by Tinkutara last updated on 20/Sep/17

The number of real solutions of the

equation 4 x^{99} + 5 x^{98} + 4 x^{97} + 5 x^{96} +

\dots . . + 4 x + 5 = 0 is

Answered by dioph last updated on 21/Sep/17

(4 x + 5) (x^{98} + x^{96} + \dots + x^{2} + 1) = 0

4 x + 5 = 0 \Rightarrow x = - \frac{5}{4}

x^{98} + x^{96} + \dots + x^{2} + 1 > 0 \forall x \in R

Hence there is only one real solution

to above equation (namely - 5 / 4)

Commented by Tinkutara last updated on 21/Sep/17

Thank you very much Sir!

Commented by dioph last updated on 21/Sep/17

Degree is even (98), and every term

is positive (+ 1) so the functions

graph never gets to cross the x axis

Commented by $@ty@m last updated on 21/Sep/17

s l i g h t m i s t a k e h e r e

I t s h o u l d b e :

(4 x + 5) (x^{98} + x^{96} + x^{94} \dots + x^{2} + 1) = 0

Commented by dioph last updated on 21/Sep/17

indeed, thanks for pointing it out .

i am most sorry and it is now corrected

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