Question

If the coefficients of the quadratic equation $a{x}^2+bx+c=0$ are odd integers, then the roots of the equation cannot be

• A. Rational number
• B. Irrational Number
• C. Imaginary number
• D. We can't say anything about the roots

Answer 1

The correct option is A Rational number

Let the roots are rational,
$a=2l+1,b=2m+1,andc=2n+1$.
then, $D={(2m+1)}^2-4(2l+1)(2n+1)$
$={odd}^2-4\times odd\times odd={odd}^2-even=say{(2p+1)}^2$
$\Rightarrow {(2m+1)}^2-4(2l+1)(2n+1)={(2p+1)}^2$ ($\because$ we assumed the roots are rational)
$\Rightarrow {(2m+1)}^2-{(2p+1)}^2=4(2l+1)(2n+1)=even$
$\Rightarrow (m-p)(m+p+1)=(2l+1)(2n+1)$
Case 1) $m$ is odd $p$ is even
LHS $=$ odd $\times$ even = even,
RHS $=$ odd, (not possible)
Similarly, for other cases: $m$ even $p$ odd
$m$ even $p$ even
$m$ odd $p$ odd
LHS is not equal to RHS.
Hence, roots cannot be rational.