How Cheenta works to ensure student success?

Explore the Back-StoryThis is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

If the coefficients of a quadratic equation , are all odd integers, show that the roots cannot be rational.

Suppose m and n are the roots. Then .

Hence if one of them is rational then the other one is rational as well.

If possible say both roots are rational. Then

where p,q,r,s are integers and g.c.d. (p,q) = 1, g.c.d.(r, s) = 1

Also .

Note that both r, s cannot be even (as their g.c.d. is 1). If possible say s is even, then r is odd. Also if s is even, p has to be even (since the 2's in s must be canceled out, as the product of the two fractions ultimately is which has odd in the numerator and odd in the denominator). Since p is even, therefore, q must be odd.

So here are the cases:

Case 1: p is even (say = 2p', q is odd, r is odd, s is even (say = 2s') (and other similar cases where crosswise numbers are even and other pair being odd)

Then .

This implies .

Or .

Since qr is odd and 4s'p' is even their sum is odd. But the denominator (2s'p) is even. Hence the numerator cannot cancel the 2's in the denominator. But that is a contradiction as this fraction is supposed to be equal to where denominator is odd.

Case 2: all of p,q,r,s are odd

.

Here the numerator is ps+qr which is even (since p, q, r, s are odd). But the denominator qs is odd. Hence 2's in the numerator won't cancel out. Hence contradiction.

Thus the roots cannot be rational.

**What is this topic:**Theory of Equations**What are some of the associated concepts:**Sum and product of roots, Vieta's formula**Book Suggestions:**Theory of Equations by Burnside and Panton

This is a Test of Mathematics Solution (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

If the coefficients of a quadratic equation , are all odd integers, show that the roots cannot be rational.

Suppose m and n are the roots. Then .

Hence if one of them is rational then the other one is rational as well.

If possible say both roots are rational. Then

where p,q,r,s are integers and g.c.d. (p,q) = 1, g.c.d.(r, s) = 1

Also .

Note that both r, s cannot be even (as their g.c.d. is 1). If possible say s is even, then r is odd. Also if s is even, p has to be even (since the 2's in s must be canceled out, as the product of the two fractions ultimately is which has odd in the numerator and odd in the denominator). Since p is even, therefore, q must be odd.

So here are the cases:

Case 1: p is even (say = 2p', q is odd, r is odd, s is even (say = 2s') (and other similar cases where crosswise numbers are even and other pair being odd)

Then .

This implies .

Or .

Since qr is odd and 4s'p' is even their sum is odd. But the denominator (2s'p) is even. Hence the numerator cannot cancel the 2's in the denominator. But that is a contradiction as this fraction is supposed to be equal to where denominator is odd.

Case 2: all of p,q,r,s are odd

.

Here the numerator is ps+qr which is even (since p, q, r, s are odd). But the denominator qs is odd. Hence 2's in the numerator won't cancel out. Hence contradiction.

Thus the roots cannot be rational.

**What is this topic:**Theory of Equations**What are some of the associated concepts:**Sum and product of roots, Vieta's formula**Book Suggestions:**Theory of Equations by Burnside and Panton

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Online Live Classroom Programs

Online Self Paced Programs [*New]

Past Papers

More