The roots of quadratic polynomials can be nice, integer values. For example \(x^2+4x+3\) has \(x=-3\) as a root. However, this is not always the case. You will have encountered many quadratic polynomials with roots that are fractions or even irrational numbers.
Convince yourself that \(x=\sqrt{2}\) is a root of the quadratic equation \(x^2-2=0\) and that \(x=\sqrt{3}\) is a root of the quadratic equation \(x^2-3=0\).
Can you find a quadratic polynomial with integer coefficients which has \(x=1+\sqrt{2}\) as a root?
What is the other root of this polynomial?
What if instead \(x=1+\sqrt{3}\) is a root? What would the quadratic polynomial be now?
What would the other root be this time?
Can you generalise your answers to the case where \(1+\sqrt{n}\) is a root?
What about the case where \(m+\sqrt{n}\) is a root?
If you think you need to revisit methods of solving quadratic equations you may like to look at these resources: Quadratic grids and Factorisable quadratics.
