Question

\(\large \color{black}{\begin{align} &\text{Find the number of integer values } \hspace{.33em}\\~\\

&\dfrac{15x^2+2x+1}{x^2-2x-1}\hspace{.33em}\\~\\

&\text{doesn't satisfy } \hspace{.33em}\\~\\
\end{align}}\)

Answer 1

Hi math,

Does that mean the number of integer values, which can not be written in the form:\[\dfrac{15x^2+2x+1}{x^2-2x-1}\]?

Answer 2

i mean the number of integer values the expression \(\dfrac{15x^2+2x+1}{x^2-2x-1}\)

not take

Answer 3

aha that's right, so we just need to find the range of function

Answer 4

\[y=\dfrac{15x^2+2x+1}{x^2-2x-1}\]find \(x\) in terms of \(y\), find the domain of inverse

Answer 5

@mukushla exact !

Answer 6

what u mean by x interms of y

Answer 7

solve the equation . and deal with y as a parameter

Answer 8

Separate x i.e. find x = ? and replace y with x afterwards.

Answer 9

u mean this

http://www.wolframalpha.com/input/?i=solve+y%3D%5Cdfrac%7B15x%5E2%2B2x%2B1%7D%7Bx%5E2-2x-1%7D+for+x

Answer 10

Following mukushla's hint
\[\large y=\dfrac{15x^2+2x+1}{x^2-2x-1}\]

cross multiplying and rearranging gives
\[\large (15-y)\color{red}{x^2}+(2+2y)\color{red}{x}+y+1 = 0\]

This is a quadratic in \(\color{red}{x}\)
you may use the discriminant to find the domain

Answer 11

ok, d=8(y+1)(y-7)

Answer 12

how?

Answer 13

from this -> (2+2y)^2-4(15-y)(y+1)

Answer 14

Ohk.. Notice that \(D \ge 0\) for the quadratic equation to make sense in real numbers

Answer 15

\(y>=7 ,y<=-1\)

Answer 16

those are the values the given rational expression ever takes

Answer 17

is infinite numbers answer

Answer 18

answer should be {0,1,2,3,4,5,6}

Answer 19

no but these numbers satisfy the equation

Answer 20

oh ok thnx

Answer 21

what on earth the question says does not satisfy??