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Given odd integers $a,b,c$ prove that the equation $ax^2 + bx + c = 0$ cannot have a solution x which is a rational number

My approach: if $a =1$ , $b=3$ , $c=1$; then the eq. $x^2 + 3x + c = 0$ has roots

$x_1=-0.381...$,

$x_2=-2.62...$ ;

which is indeed true but how can we prove this result?? please help

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Suppose $a,b,c$ are odd integers, and suppose the equation $$ax^2+bx+c=0$$ has a rational root, $r={\large{\frac{p}{q}}}$ say, where $p,q$ are integers, and $\text{gcd}(p,q)=1$ (i.e., the fraction is reduced to lowest terms). \begin{align*} \text{Then}\;\;&ar^2 + br + c = 0\\[4pt] \implies\;&a\left({\small{\frac{p}{q}}}\right)^2+b\left({\small{\frac{p}{q}}}\right) + c = 0\\[4pt]\implies\;&ap^2 + bpq + cq^2 = 0\\[4pt] \end{align*} Consider $3$ cases . . .

Case $(1)\,$:$\;p,q$ are both even.

Then $\text{gcd}(p,q) > 1$, contradiction.

Case $(2)\,$:$\;p,q$ are both odd.

Then the LHS of the equation $ap^2 + bpq + cq^2 = 0$ is odd (since all three terms are odd), contradiction, since the RHS is even.

Case $(3)\,$:$\;$One of $p,q$ is even, and the other is odd.

Then the LHS of the equation $ap^2 + bpq + cq^2=0$ is odd (since two of the terms are even, and the remaining term is odd), contradiction, since the RHS is even.

Thus, all $3$ cases yield a contradiction.

It follows that the equation $ax^2 + bx + c = 0$ does not have a rational root.

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If $c$ is integer and the roots are rational then both roots have an odd numerator (since their product is $c/a$). Can you finish?

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We need $$b^2-4ca$$ for the perfect square $=d^2$(say)

If $b$ is odd, so will be $d$

$$\implies4ca=b^2-d^2$$

Now if $x$ is odd, $$x^2\equiv1\pmod8\implies8|(b^2-d^2)$$ but $8\nmid4ca$

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