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let $P(x)$ and $Q(x)$ be 2 quadratic eqs. with one non-rational root common and integral coefficients. prove $P(x) = r.Q(x)$, for some rational no. $r$

TRIED ANSWER:

Let $P(x)= ax^2 + bx + c$

$\Rightarrow$ $x = \frac{-b \pm \surd b^2-4ac}{2a}$

Let $Q(x)= mx^2 + nx + o$

$\Rightarrow$ $x' = \frac{-n \pm \surd n^2-4mo}{2m}$

If one root of $P(x)$ = one root of $Q(x)$

then assume $\Rightarrow$ $\frac{-b + \surd b^2-4ac}{2a}$ = $\frac{-n + \surd n^2-4mo}{2m}$

As they both are irrational, comparing the two we get the irrational part equal.

Hence both the roots are same for the two equations

OR

They are the same equations with different leading coefficients.

THUS

$P(x)=k*Q(x)$

for some constant k.

Alok Ranjan
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  • Do you mean polynomials? What have you tried? – Daniel R May 28 '14 at 09:12
  • @AlokRanjan: You didn't answer Daniel R. – Michael Albanese May 28 '14 at 09:27
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    @AlokRanjan You will receive better quality answers if you include your own working up to the point you got stuck. Then posters have a better idea of where you're going wrong and can point you in the right direction. – Graham Kemp May 28 '14 at 09:29
  • For the statement to be true, I think that it need to be assumed that $P$ and $Q$ have rational coefficients. – ajotatxe May 28 '14 at 09:49
  • @MichaelAlbanese Why is the question on hold?? – Alok Ranjan Jun 01 '14 at 10:27
  • @AlokRanjan: The reason your question was put on hold is written below. Since you edited the question, it has been put in the reopen queue, but it takes time for it to be reopened (if it should be) because it takes five votes from people with at least $3,000$ votes. – Michael Albanese Jun 01 '14 at 10:30

2 Answers2

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The statement is false. For example, $P(x)=(x-\sqrt 2) = x^2 - 2\sqrt 2 x + 2$ and $Q(x) = x^2-2$ have one non-rational root in common. However, there is no rational number $r$ such that $P(x) = r\cdot Q(x)$.

5xum
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Let $P(x)=ax^2+bx+c$ and $Q(x) = dx^2+ex+f$. Let the roots of $P(x)$ be $\alpha,\beta$ and of $Q(x) $ be $\alpha,\gamma$. $\alpha$ is the common irrational root.
$\alpha+\beta=\frac{-b}a$. Since $-\frac ba$ is irrational, $\alpha+\beta$ must also be irrational. $\implies\beta=-\alpha$. $\implies -\frac ba=0 \implies b = 0$
Similarly, $\gamma = -\alpha$ and $e=0$.

$\alpha\cdot\beta=\frac ca \implies-\alpha^2=\frac ca = \frac fd$.

$P(x)=ax^2+c$ and $Q(x)=dx^2+f$.
$P(x)=ax^2-a\cdot\alpha^2$ and $Q(x)=dx^2-d\alpha^2$.
$P(x)=ax^2-a\cdot\alpha^2=\frac ad(dx^2-d\cdot\alpha^2)=\frac adQ(x)=r\cdot Q(x)$.

Shaurya Gupta
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