Rational & Real Roots in Quadratic Equations: Explained

[Drain Brain]

Can a quadratic equation with rational coefficients have one rational root and one irrational root? explain.

and

Can a quadratic equation with real coefficients have one real
root and one imaginary root? Explain.

please enlighten me.

 

[MarkFL]

I would begin by looking at the quadratic formula, the discriminant in particular. What do you find?

 

[Drain Brain]

I would begin by looking at the quadratic formula, the discriminant in particular. What do you find?

If the discriminant $b^2-4ac>0$ the roots are both rational and cannot be rational and irrational at the same time. Am I making sense here?

 

[soroban]

Hello, Drain Brain!

If the discriminant $b^2-4ac>0$, the roots are both rational,
and cannot be rational and irrational at the same time.
Am I making sense here?

You're close . . .

If the discriminant $b^2-4ac$ is a square,
. . the roots are both rational.

If the discriminant $b^2-4ac$ is positive,
. . the roots are both real.

If the discriminant $b^2-4ac$ is negative,
. . the roots are both complex (imaginary).

We can never have "one of each".

 

[MarkFL]

If the discriminant $b^2-4ac>0$ the roots are both rational and cannot be rational and irrational at the same time. Am I making sense here?

Yes, if one root is rational, then we know $$\sqrt{b^2-4ac}$$ must itself be rational, and therefore both roots must be rational. So, given $a$, $b$ and $$\sqrt{b^2-4ac}$$ are all rational, we may express the roots as:

$$\frac{\dfrac{p_1}{q_1}\pm\dfrac{p_2}{q_2}}{\dfrac{p_3}{q_3}}=\frac{q_3}{p_3}\left(\frac{p_1}{q_1}\pm\frac{p_2}{q_2}\right)=\frac{q_3}{p_3}\left(\frac{p_1q_2\pm p_2q_1}{q_1q_2}\right)=\frac{p_1q_2q_3\pm p_2q_1q_3}{p_3q_1q_2}$$

Can you see how both must be rational?

 

[Drain Brain]

I kind of understand this

$\frac{p_1}{q_1}$, $\frac{p_2}{q_2}$ and $\frac{p_3}{q_3}$ are representation of rational numbers right?

 

[MarkFL]

I kind of understand this

$\frac{p_1}{q_1}$, $\frac{p_2}{q_2}$ and $\frac{p_3}{q_3}$ are representation of rational numbers right?

Yes, that right, and they are assumed to be in fully reduced form, that is $p_i$ and $q_i$ have no common factors. :D

So, in the final form I gave above, we have an integer in the denominator, and the sum/difference of integers in the numerator, which will be integers. So, the two roots are integers divided by integers, which are by definition, rational numbers.