Is is possible for a quadratic equation with only one irrational root to have integral coefficients?

Question

Given a quadratic equation with one and only one root (for example $6-\sqrt{2}$ ). Does there exist integers $a,b$ and $c$ where $ax^2 + bx + c = 0$ for that root?

Answer 1

Not even if the coefficients are rational. If a quadratic equation has $r$ and $s$ as roots, it can be written as $(x-r)(x-s) =x^2-(r+s)x + rs $.

If the coefficients are rational, then $r+s$ is rational, so $r$ and $s$ are both rational or both irrational.

Answer 2

Yes. given $r_1 = 6-\sqrt{2}$, we know that $r_2 = 6+\sqrt{2}$

so the quadratic equation is given by

$(x-(6-\sqrt{2}))(x-(6+\sqrt{2}))=(x-6)^2-2 = x^2-12x+34$

Answer 3

When there is only one root, the discriminant is zero and the root is $\frac{-b}{2a}$ because $\frac{-b+0}{2a}$ and $\frac{-b-0}{2a}$ are both equal to $\frac{-b}{2a}$

It is impossible for a quadratic equation with integer coefficients to have only one irrational root because $\frac{-b}{2a}$ will always be rational when the coefficients are integers. That's simply because a rational number is defined as a number that can be expressed as a ratio of two integers.