Solve the given inequality Q. 21

AI Thread Summary
The inequality discussed is 5x² - cx² - 4x - 2 < 0, leading to the condition 56 - 8c < 0, which simplifies to c > 7. However, the textbook states 7 > c, creating confusion. The user realizes they made an error by using the wrong inequality sign and corrects it to b² - 4ac > 0. They confirm that for the inequality to hold, 5 - c must be less than 0, ensuring the x² term does not approach infinity. The discussion concludes with an acknowledgment of the mistake and a friendly note to a fellow participant.
chwala
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Homework Statement
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Relevant Equations
Discriminant
Q. 21

Allow me to post using phone... Will amend/ show working when i get hold of computer.In my working I have
##5x^2 -cx^2-4x-2<0##
##56-8c<0##

...

##c>7##

Textbook says ##7>c## ...who's fooling who here? 😊

I have seen my mistake. I am wrong...put in wrong inequality sign...
Should be ##b^2 - 4ac >0 ##

Cheers guys
 

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(5-c)x^2-4x-2&lt;0
We know that 5-c<0 at least so that x^2 term does not go plus infinity as |x| goes infinity.
 
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Likes e_jane and chwala
anuttarasammyak said:
(5-c)x^2-4x-2&lt;0
We know that 5-c<0 at least so that x^2 term does not go plus infinity as |x| goes infinity.
I saw my mistake. My bad. Hope you good @anuttarasammyak been with you here for some time mate ...
 

Thread 'Greatest possible value of a constant in polynomial'

Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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