ReducFind the illegal value of b
- Thread starter grpriest
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- Feb 4, 2004
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What you have posted means the following:
. . . . .\(\displaystyle \L 2b^2\,+\,3b\,-\,\frac{10}{b^2}\,-\,2b\,-\,8\)
Is this what you mean? Or do you mean "(2b<sup>2</sup> + 3b - 10)/(b<sup>2</sup> - 2b - 8)", which may also be written as the following?
. . . . .\(\displaystyle \L \frac{2b^2\,+\,3b\,-\,10}{b^2\,-\,2b\,-\,8}\)
By "illegal values", do you mean "values which would make the fraction undefined?
What have you tried on this exercise so far? (For instance, you could try plugging in the solution options, to see which works.) How far have you gotten? Where are you stuck?
Thank you.
Eliz.
. . . . .\(\displaystyle \L 2b^2\,+\,3b\,-\,\frac{10}{b^2}\,-\,2b\,-\,8\)
Is this what you mean? Or do you mean "(2b<sup>2</sup> + 3b - 10)/(b<sup>2</sup> - 2b - 8)", which may also be written as the following?
. . . . .\(\displaystyle \L \frac{2b^2\,+\,3b\,-\,10}{b^2\,-\,2b\,-\,8}\)
By "illegal values", do you mean "values which would make the fraction undefined?
What have you tried on this exercise so far? (For instance, you could try plugging in the solution options, to see which works.) How far have you gotten? Where are you stuck?
Thank you.
Eliz.