isi-entrance 2006 Q2

isi-entrance · India · solved Discriminant and conditions for roots Nature of roots given coefficient constraints

a) Show that if $a$ and $b$ are irrational numbers that are roots of a quadratic with rational coefficients, then $(a-b)^2$ is not a perfect square of any rational number.
b) i) If $a = r \pm \sqrt{s}$ is a quadratic surd, find a rational $x$ such that $a + x$ is irrational but $a_n = (r + (r^2 - s)) \pm \sqrt{s} \notin \mathbb{Q}$. If $a$ is not a surd, take $x = -a$.
ii) Find $y$ such that the required condition holds.

a) Then $a$ and $b$ are the roots of the quadratic with rational coefficients $x^2 - (a+b)x + ab = 0$ then $a = \frac{a+b}{2} \pm \frac{\sqrt{\Delta}}{2}$ where $\Delta = (a-b)^2$. Clearly, $(a-b)^2 \neq c^2$ for any rational $c$, since that would lead to $a \in \mathbb{Q}$.
b) i) If $a = r \pm \sqrt{s}$ is a quadratic surd, take $1 + r \mp \sqrt{s}$ then $x = 1 + 2r \in \mathbb{Q}$ and $a_n = (r + (r^2 - s)) \pm \sqrt{s} \notin \mathbb{Q}$. If $a$ is not a surd, take $x = -a$.
ii) Take $y = 0$.

a) Show that if $a$ and $b$ are irrational numbers that are roots of a quadratic with rational coefficients, then $(a-b)^2$ is not a perfect square of any rational number.

b) i) If $a = r \pm \sqrt{s}$ is a quadratic surd, find a rational $x$ such that $a + x$ is irrational but $a_n = (r + (r^2 - s)) \pm \sqrt{s} \notin \mathbb{Q}$. If $a$ is not a surd, take $x = -a$.

ii) Find $y$ such that the required condition holds.

Paper Questions

Q1 Q2 Q3 Q4 Q5 Q7 Q8 Q9 Q10