Question

If a, b, c are odd integers and $a{x}^2+bx+c=0$ , has real roots then:

• A. Both roots are rational
• B. Both roots are irrational
• C. Both roots are positive
• D. Roots are of opposite signs.

Answer 1

The correct option is B

Both roots are irrational

a, b, c are odd integers, $D=b^2-4ac\ge 0$

If equation has rational roots, then $b^2-4ac$ is perfect square. Since $b^2-4ac$ (odd - even) is odd, $b^2-4ac$ is square of odd integer.

Let $b^2-4ac$ = $(2k+1{)}^2$

$\Rightarrow 4ac$ = $(2n+1{)}^2$ - $(2k+1{)}^2$ [ b = 2n + 1 ]

=(2n + 1 - 2k - 1) (2n + 2k + 2)

=4 (n - k) (n + k + 1)

$\Rightarrow$(n - k)(n + k + 1) = ac

$\Rightarrow$even interger = odd integer which is not true.

Hence, both roots must be irrational.