Rational or Irrational number

Question

we know that "$a$" is a Irrational number .But "$a^2+a$" is Rational. Can You find "$a$"? (more than one answer is available)

Answer 1

We have $a^2+a=r\in\Bbb Q$ if and only if $$a=\frac{-1\pm\sqrt{1+4r}}{2}.$$ Any $r\in\Bbb Q$ such that $1+4r$ is not a perfect square will do.

Answer 2

Hint complete the square you get $(a+\frac{1}{2})^2-\frac{1}{4}$ so any number of the form $√n-\frac{1}{2}$ will give it as a cancels out with $-a$ making number rational.

Answer 3

Consider $x^2 + x = \dfrac pq$
for some relatively prime integers $p$ and $q$.

Hence $qx^2 + qx - p = 0$

So, for any relatively prime integers $p$ and $q$ such that $q^2 + 4pq$ is positive and not a perfect square,

$a \in \left\{ \dfrac{-q \pm \sqrt{q^2 + 4pq}}{2q} \right\}$

will have the property that $a$ is irrational and $a^2 + a = \dfrac pq$.