First slide

Theory of equations

Question

If the roots of the equation $(p^{2} + q^{2}) x^{2} - 2 q (p + r) x + (q^{2} + r^{2}) = 0$ be real and equal, then p, q, r will be in

Moderate

A

A.P.
B

G.P.
C

H.P.
D

None of these

Solution

Given equation is $4 q^{2} {(p + r)}^{2} - 4 (p^{2} + q^{2}) (q^{2} + r^{2}) = 0$
Roots are real and equal, then
$4 q^{2} {(p + r)}^{2} - 4 (p^{2} + q^{2}) (q^{2} + r^{2}) = 0 \Rightarrow q^{2} (p^{2} + r^{2} + 2 pr) - (p^{2} q^{2} + p^{2} r^{2} + q^{4} + q^{2} r^{2}) = 0 \Rightarrow q^{2} p^{2} + q^{2} r^{2} + 2 {pq}^{2} r - p^{2} q^{2} - p^{2} r^{2} - q^{4} - q^{2} r^{2} = 0 \Rightarrow 2 {pq}^{2} r - p^{2} r^{2} - q^{4} = 0 \Rightarrow {(q^{2} - pr)}^{2} = 0 Hence q^{2} = pr . Thus p, q, r in G . P .$

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