1

The question is to find out the value of $a$ such that $ax^2-4x+9=0$ has integer roots.

If $a$ is 1 then the roots if they are rational are integers.But the discriminant in such case is -ve so $a\neq 1$ Similarly by rational root theorem $a \neq -1$ So $a$ must be a fraction of the form $1/b$ such that $b$ is a integer. Now the condition of the equation $x^2-4bx+9b$ to have rational roots and thereby integers because the leading coefficient is unity is that its discriminant must be a perfect square.so $16b^2-36b$ is a perfect square.I couldn't get how to determine $b$ from this.Any ideas?Thanks

This link Elementary number theory and quadratic doesn't have the attempt in which I am unable to proceed.

user471651
  • 1,164
  • Must both roots be rational, or only one root? – Toby Mak Dec 15 '17 at 13:42
  • @TobbyMark both roots must be rational so that since the leading coefficient is unity both roots become integers which is what the question states. – user471651 Dec 15 '17 at 13:51
  • 2
    https://www.quora.com/What-is-the-value-of-a-if-ax-2-4x+9-0-has-integral-roots –  Dec 15 '17 at 13:53
  • @kingW3 no the wording of the question may be difference but the trouble I am facing in the question is not discussed there. – user471651 Dec 15 '17 at 14:29

1 Answers1

6

Starting from your original equation: $$ ax^2 -4x+9 = 0\text{, immediately } x = \frac{2\pm\sqrt{4-9a}}{a} $$ Insisting that $x\in\Bbb{Z}$ gives $4-9a = n^2$ for some $n\in\Bbb{Q}$ and so $$ \frac{9(2\pm n)}{4-n^2} \in \Bbb{Z}\quad \frac{9}{2+ n}\in \Bbb{Z}\text{ and }\frac{9}{2- n}\in \Bbb{Z} $$ There are finitely many $n$ in $\Bbb{Q}$ that satisfy the two conditions simultaneously. If you insist on only one integral root then just choose $n$ so that $\frac{9}{2+ n}\in \Bbb{Z}$ and solve for $a$.

Malcolm
  • 1,055