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If $p, q$ are odd integers, then the roots of the equation $2 p x^{2}+(2 p+q) x+q=0$ are
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Verified Answer
The correct answer is:
rational
Given equation.
$$
\begin{array}{l}
2 p x^{2}+(2 p+q) x+q=0 \\
\therefore D=(2 p+q)^{2}-4(2 p)(q) \\
\quad=4 p^{2}+q^{2}+4 p q-8 p q \\
\quad=4 p^{2}+q^{2}-4 p q \\
\quad=(2 p-q)^{2}
\end{array}
$$
$=$ a perfect square
Given equation has rational roots
$$
\begin{array}{l}
2 p x^{2}+(2 p+q) x+q=0 \\
\therefore D=(2 p+q)^{2}-4(2 p)(q) \\
\quad=4 p^{2}+q^{2}+4 p q-8 p q \\
\quad=4 p^{2}+q^{2}-4 p q \\
\quad=(2 p-q)^{2}
\end{array}
$$
$=$ a perfect square
Given equation has rational roots
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