If a, b, c are odd integers and ax2 + bx + c = 0, has real roots then :

a,b,c are odd integers, D = b2 - 4ac > 0.If equation has rational roots, the n b2 - 4ac isperfect square. Since b2 - 4ac (odd-even) is odd, b2 - 4ac is square of odd integer.Let b2−4ac=(2k+1)2⇒4ac=(2n+1)2−(2k+1)2[b=2n+1]=(2n+1−2k−1)(2n+2k+2)=4(n−k)(n+k+1)⇒(n−k)(n+k+1)=ac⇒even integer = odd integerwhich is not true.Hence, both roots must be irrational.

If a, b, c are odd integers and ax2 + bx + c = 0, has real roots then :

a,b,c are odd integers, D = b2 - 4ac > 0.If equation has rational roots, the n b2 - 4ac isperfect square. Since b2 - 4ac (odd-even) is odd, b2 - 4ac is square of odd integer.Let b2−4ac=(2k+1)2⇒4ac=(2n+1)2−(2k+1)2[b=2n+1]=(2n+1−2k−1)(2n+2k+2)=4(n−k)(n+k+1)⇒(n−k)(n+k+1)=ac⇒even integer = odd integerwhich is not true.Hence, both roots must be irrational.

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If a, b, c are odd integers and ax² + bx + c = 0, has real roots then :

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both roots are rational

roots are of opposite signs

both roots are irrational

both roots are positive

answer is B.

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Detailed Solution

a,b,c are odd integers, D = b² - 4ac > 0.
If equation has rational roots, the n b² - 4ac is
perfect square. Since b² - 4ac (odd-even) is odd, b² - 4ac is square of odd integer.
$\begin{array}{l} L e t b^{2} - 4 a c = (2 k + 1)^{2} \\ \Rightarrow 4 a c = (2 n + 1)^{2} - (2 k + 1)^{2} [b = 2 n + 1] \\ = (2 n + 1 - 2 k - 1) (2 n + 2 k + 2) \\ = 4 (n - k) (n + k + 1) \\ \Rightarrow (n - k) (n + k + 1) = a c \end{array}$
$\Rightarrow$ even integer = odd integer
which is not true.
Hence, both roots must be irrational.