Step by step solution to "If the coefficients of a quadratic equation are..."

abiejose55d

Answered question

2022-04-23

If the coefficients of a quadratic equation are odd numbers, show that it cannot have rational roots

a x^{2} + b x + c = 0

eslasadanv3

Beginner2022-04-24Added 20 answers

Step 1
Let the quadratic be

f (x) = a x^{2} + b x + c

where

a, b, c \equiv 1 \pm o d {2}

By the Rational Root Theorem, if

\frac{p}{q}

is a root of the quadratic in its lowest terms, then

p ∣ c

and

q ∣ a

Since a and c are odd, then both p and q must be odd. Then, we have

f (\frac{p}{q}) = a \cdot \frac{p^{2}}{q^{2}} + b \cdot \frac{p}{q} + c = \frac{a p^{2} + b p q + c q^{2}}{q^{2}} .

However, we have that a,b,c,p, and q are all odd, so then

a p^{2} + b p q + c q^{2}

is also odd, which means we cannot have

f (\frac{p}{q})

by contradiction. Therefore, the quadratic f(x) cannot have any rational roots.

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