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With the quadratic $x^2-6x-5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

If one root is irrational, then $\Delta = \sqrt{b^2-4ac}$ in the quadratic formula must be irrational. Because of the quadratic formula $x = \frac{-b ± \Delta}{2a}$, either both roots are rational or both roots are irrational. This concept is further explained in this post.

Now that we have confirmed there must be two rational roots, let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

- Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.

- Case $2$: $qr$ is odd and $ps$ is even. The same logic applies, so either one of $qs$ or $pr$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

With the quadratic $x^2-6x+5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

If one root is irrational, then $\Delta = \sqrt{b^2-4ac}$ in the quadratic formula must be irrational. Because of the quadratic formula $x = \frac{-b ± \Delta}{2a}$, either both roots are rational or both roots are irrational. This concept is further explained in this post.

Now that we have confirmed there must be two rational roots, let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

- Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.

- Case $2$: $qr$ is odd and $ps$ is even. The same logic applies, so either one of $qs$ or $pr$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

With the quadratic $x^2-6x-5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

Each of $p, q, r, s$ appears once in the terms $qs$ and $pr$. For $qs$ and $pr$ to be rational, both $q, s$ or $p, r$ must be irrational (say $\sqrt{2} \times \sqrt{2} = 2$). The only way to force $qr + ps$ to be rational as well is to make $qr = -ps$, from which we conclude that $p,q,r,s$ must all be integers (i.e that $\frac{p}{q}$ and $\frac{r}{s}$ are both rational).

Now let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

- Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.

- Case $2$: $qr$ is odd and $ps$ is even. The same logic applies, so either one of $qs$ or $pr$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

With the quadratic $x^2-6x-5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

If one root is irrational, then $\Delta = \sqrt{b^2-4ac}$ in the quadratic formula must be irrational. Because of the quadratic formula $x = \frac{-b ± \Delta}{2a}$, either both roots are rational or both roots are irrational. This concept is further explained in this post.

Now that we have confirmed there must be two rational roots, let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

- Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.

- Case $2$: $qr$ is odd and $ps$ is even. The same logic applies, so either one of $qs$ or $pr$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

With the quadratic $x^2-6x-5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

Each of $p, q, r, s$ appears once in the terms $qs$ and $pr$. For $qs$ and $pr$ to be rational, both $q, s$ or $p, r$ must be irrational (say $\sqrt{2} \times \sqrt{2} = 2$). The only way to force $qr + ps$ to be rational as well is to make $qr = -ps$, from which we conclude that $p,q,r,s$ must all be integers (i.e that $\frac{p}{q}$ and $\frac{r}{s}$ are both rational).

Now let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

• Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $q \times s$ or $p \times r$ would be even.

• Case $2$: $qr$ is odd and $ps$ is even.

The same logic applies, so either one of $q \times s$ or $p \times r$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.

With the quadratic $x^2-6x-5$, you can cross off options $B, C, D$ which leaves only option $A$ by the process of elimination. However, as you stated in the comments, this is not what you wanted.

Let $p/q$ and $r/s$ be the two roots of the quadratic. You can form a quadratic equation which satisfies these conditions as follows:

$$\left(x-\frac{p}{q} \right) \left(x - \frac{r}{s} \right) = 0$$ $$\iff \left(qx-p \right) \left(sx-r \right) = 0$$ $$\iff (qs)x^2-(qr+ps)x+pr = 0$$

Each of $p, q, r, s$ appears once in the terms $qs$ and $pr$. For $qs$ and $pr$ to be rational, both $q, s$ or $p, r$ must be irrational (say $\sqrt{2} \times \sqrt{2} = 2$). The only way to force $qr + ps$ to be rational as well is to make $qr = -ps$, from which we conclude that $p,q,r,s$ must all be integers (i.e that $\frac{p}{q}$ and $\frac{r}{s}$ are both rational).

Now let's check if condition $A$ must always be true. The contrapositive of 'at least one of $a,b,c$ is even is that none of $a,b,c$ are even, which means $qs, qr + ps$ and $pr$ must all be odd. If $qr$ and $ps$ are both even, then at least one of $qs$ and $pr$ is even. However, if $qr$ and $ps$ are both odd, then the resulting sum is also even.

This means that either $qr$ is even and $ps$ is odd, or that $qr$ is odd and $ps$ is even.

Now would be a good time to stop and continue the proof for yourself. If you are interested, the rest of the proof is hidden behind the spoiler effect:

- Case $1$: $qr$ is even and $ps$ is odd

Since $qr$ is even, at least one of $q$ and $r$ is even. However, since $\text{even} \times \text{odd} = \text{even}$, either one of $qs$ or $pr$ would be even.

- Case $2$: $qr$ is odd and $ps$ is even. The same logic applies, so either one of $qs$ or $pr$ would be even.

This means that the quadratic cannot be written in the form $(qs)x^2-(qr+ps)x+pr = 0$. Reversing the order of the steps means that the quadratic does not have two rational roots. By proving the contrapositive of the statement, we have proved that option $A$ is indeed correct.