if the roots of the equation (p^2+q^2)x² + 2(qr –ps)x+(r^2+s^2) = 0 are real and equal, prove that pr+qs=0
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Answer:
B. ps=rq
Step-by-step explanation:
∝=β
2∝
−b−
b
2
−4ac
=
2∝
−2+
b
2
−4ac
b
2
−4ac=0
[−2(pq+qs)]
2
−4(p
2
+q
2
)(r
2
+s
2
)=0
4p
2
r
2
+4q
2
s
2
+8pqrs−4p
2
r
2
−4q
2
r
2
−4q
2
s
2
p
2
s
2
+q
2
r
2
−2pqrs=0
(ps−qr)
2
=0
ps=qr
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