(ii) (a+2b)^+(a-2b)^=2(a²+4b^)
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Answer:
The roots of (a
2
+b
2
)x
2
−2b(a+c)x+(b
2
+c
2
)=0 are equal.
i.e. D=b
2
−4ac=0
⇒4b
2
(a+c)
2
=4(a
2
+b
2
)(b
2
+c
2
)
⇒b
2
(a
2
+c
2
+2ac)=a
2
b
2
+a
2
c
2
+b
4
+b
2
c
2
⇒b
2
a
2
+b
2
a
2
+2ab
2
c=a
2
b
2
+a
2
c
2
+b
4
+b
2
c
2
⇒b
4
−2b
2
ac+a
2
c
2
=(b
2
−2ac)
2
=0
∴b
2
=ac
i.e. a,b,c are in GP.
Hence, option B.
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Answer:
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