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This is from a summer work packet for high school AP Calculus. I've never seen anything like this. I played around with it and got this as a best guess: $$\{ b \in \Bbb{Q} \; | \; b \le 2.25 \}$$ I'm not sure if that's formatted the right way but whatever. Sorry in advanced if I am breaking any rules for asking questions (I just skimmed through the "How to Ask" page).

(My bad about copying the wrong problem)

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  • $\begingroup$ If a quadatic is not factorisable it means it has no real roots.... $\endgroup$ – Meadara Aug 8 '16 at 17:08
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    $\begingroup$ It depends what "factorable" means in this context. From the context, I'm guessing that the working definition is that a polynomial is factorable iff it has rational roots. $\endgroup$ – Omnomnomnom Aug 8 '16 at 17:08
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    $\begingroup$ A good idea here is to look at the discriminant $b^2 - 48$. $\endgroup$ – Omnomnomnom Aug 8 '16 at 17:09
  • $\begingroup$ What do you mean by "factorable"? Factorable over $\Bbb Z$, over $\Bbb Q$, over $\Bbb R$? $\endgroup$ – Alex M. Aug 8 '16 at 17:15
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For the expression to be factorable (over the real numbers), the quadratic equation $$ x^2 - 3x + b = 0 $$ has to have two real (not necessarily distinct) roots. This condition is equivalent to the discriminant being nonnegative: $$ 3^2 - 4b \geq 0. $$

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  • $\begingroup$ Does this mean I'm right? Thanks for the stuff about discriminates thats helpful (also @Omnomnomnom cuz i cant upvote comments yet) $\endgroup$ – Cahion3 Aug 8 '16 at 17:31
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$x^2 - 3x + b = (x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1r_2$ implies that $3 = r_1+r_2$ and $b = r_1r_2$

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  • $\begingroup$ I'm confused sorry $\endgroup$ – Cahion3 Aug 8 '16 at 17:31
  • $\begingroup$ @Cahion3 It means you can pick whatever value you want for $r_1$ and from there you automatically get $r_2$ and $b$. There are infinitely many solutions, unless there is extra wording in your problem putting any restrictions on anything. $\endgroup$ – AJJ Aug 8 '16 at 18:00

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