If $p, q$ are real and $p \neq q$ then show that the root of the equation $(p - q) x^{2} + 5 (p + q) x - 2 (p - q) = 0$ are real and unequal.

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Solution

$\begin{matrix} (p - q) x^{2} + 5 (p + q) x - 2 (p - q) = 0 H e r e, a = p - q b = 5 (p + q) c = - 2 (p - q) A s t h e r o o t s a r e r e a l a n d u n e q u a l, b^{2} - 4 a c > 0 i e . {5 (p + q)}^{2} - 4 (p - q) (- 2 {p - q}) 25 (p^{2} + 2 p q + q^{2}) + 8 (p - q) (p - q) 25 (p^{2} + 2 p q + q^{2}) + 8 {(p - q)}^{2} 25 (p^{2} + 2 p q + q^{2}) + 8 p^{2} - 16 p q + 8 q^{2} 25 p^{2} + 50 p q + 25 q^{2} + 8 p^{2} - 16 p q + 8 q^{2} > 0 \end{matrix}$

Hence, the equation has real and unequal roots.

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