Let's understand step by step:
- A quadratic equation is in the form: \(ax^2 + bx + c = 0\), where \(a, b, c\) are called coefficients.
- Coefficients can be any real numbers: they may be rational (like \(2, -3, \dfrac{1}{2}\)) or irrational (like \(\sqrt{2}, \pi\)).
- We want to check: is it possible that the coefficients are irrational but the solutions (roots) are rational numbers?
Example:
- Take the equation: \(\sqrt{2}x^2 - \sqrt{2} = 0\).
- Here, the coefficient of \(x^2\) is \(\sqrt{2}\) (irrational) and the constant term is \(-\sqrt{2}\) (irrational).
Now solve it:
- \(\sqrt{2}x^2 - \sqrt{2} = 0\)
- Take out \(\sqrt{2}\) as common: \(\sqrt{2}(x^2 - 1) = 0\)
- So, \(x^2 - 1 = 0\)
- Which gives: \(x^2 = 1\)
- Therefore, \(x = \pm 1\)
Thus, the roots are \(1\) and \(-1\), both rational numbers.
Conclusion: Yes, it is possible. The equation \(\sqrt{2}x^2 - \sqrt{2} = 0\) has irrational coefficients but rational roots.