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Question
If p, q are real and p ≠ q, then show that the roots of the equation (p − q) x2 + 5(p + q) x− 2(p − q) = 0 are real and unequal.
Solution
The quadric equation is (p − q) x2 + 5(p + q) x− 2(p − q) = 0
Here,
a = (p - q), b = 5(p + q) and c = -2(p - q)
As we know that D = b2 - 4ac
Putting the value of a = (p - q), b = 5(p + q) and c = -2(p - q)
D = {5(p + q)}2 - 4 x (p - q) x (-2(p - q))
= 25(p2 + 2pq + q2) + 8(p2 - 2pq + q2)
= 25p2 + 50pq + 25q2 + 8p2 - 16pq + 8q2
= 33p2 + 34pq + 33q2
Since, P and q are real and p ≠ q, therefore, the value of D ≥ 0.
Thus, the roots of the given equation are real and unequal.
Hence, proved
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