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This is what I've done so far:

Using the formula, we get

$x = \frac{-c \pm \sqrt{c^2 - 24}}{2}$

For solutions to exist, $c \geq 5$. For solutions to be rational, $c^2 - 24$ has to be a perfect square.

I'm kind of stuck now. I'd appreciate if someone can give me a hint or two on how to solve this question. Thanks!

OinkOink
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3 Answers3

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If the solutions of $x^2+cx+6$ are rational, then $$ x^2+cx+6=(x-a)(x-b)=x^2-(a+b)x+ab $$ for rational $a,b$. Then we have $ab=6$, and $c$ is given by $c=-a-b$. We can choose any rational number $a\neq 0$, then $b=6/a$ and $c=-\frac{a^2+6}{a}$.

Dietrich Burde
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  • Is there a simpler way of solving the question? This is actually a high school math question and my student has yet to learn that. – OinkOink Aug 28 '16 at 07:49
  • I find a square root more difficult than multiplying out $(x-a)(x-b)$. What do you mean then by "simpler" ? – Dietrich Burde Aug 28 '16 at 07:50
  • Is there an alternative method? I can't explain the question to my student using the method that you've suggested because he has yet to learn the concept of the sum and product of roots. – OinkOink Aug 28 '16 at 07:56
  • We do not need the concept of the sum and product of roots. We just need comparison of coefficients, and that $(x-a)(x-b)=0$ means that all solutions of $x^2-(a+b)+ab=0$ are given by $a$ and $b$. – Dietrich Burde Aug 28 '16 at 08:15
  • Kay, I get what you mean now. Thanks a lot! :) – OinkOink Aug 28 '16 at 08:17
  • You are welcome! – Dietrich Burde Aug 29 '16 at 07:31
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Observe that if $x \in \mathbb{Q}$ which is a solution of the equation $x^2 + cx + 6 = 0\text{ } (1)$, then solve for $c$ in terms of $x$ we obtain: $c = -\dfrac{x^2+6}{x}$. Thus if $c = - \dfrac{r^2+6}{r}$ with $r \in \mathbb{Q}\setminus \{0\}$, we prove the solutions of $(1)$ are rational numbers. We have: $x^2+ cx + 6 = 0 \implies x^2 - \left(\dfrac{r^2+6}{r}\right)x + 6 = 0\implies x^2r-r^2x - 6(x-r) = 0\implies (xr-6)(x-r)=0\implies x = r, \dfrac{6}{r}$ which are both rationals.

DeepSea
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0

1) What is a rational number ? It is a number which can be written in the form $\frac{p}{q} $ , where $q \neq 0$ and $p , q \in \mathbb Z$

2)

In this case : Let's take the quadratic equation $ax^2+bx+c=0$ where $a \neq 0$.

We all know that the roots are given by , $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

Case 1 : Suppose that $a,b,c \in \mathbb Q$.

Then $x$ is rational if and only if $b^2-4ac$ is a perfect square or zero $(0,1,4,9,16,...)$.

Now we need to make $b^2-4ac$ a perfect square !

Now observe that if $a+b+c=0$ , then $b^2-4ac = (-a-c)^2-4ac=(a-c)^2$

That is if $a,b,c \in \mathbb Q$ and $a+b+c=0$ then the solutions are rational.

Case 2 : Suppose that $a,b,c \in \mathbb Q$.

If $c=0$ , then all the roots are rational. (This is easy if all $a,b,c$ are rationals)

Case 3 : Suppose that $b,c \in \mathbb R- \mathbb Q$.(If $a$ is irrational we can divide by $a$ )

$a+b+c=0$ condition does not satisfy.

Ex : $(1-\sqrt{2})x^2-2x+(1+\sqrt{2})=0$

Angelo Mark
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