Let a, b, c be real numbers a≠0. If p is a root of a2y2+by + c = 0, q is the root of a2y2+by + c = 0 and 0 <p< q, then the equation a2y2+by + c = 0 has a root r that always satisfies
(a) r = (p+q)/2
(b) r = p + (q/2)
(c) r = a
(d) p <r <q
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Let a, b, c be real numbers a≠0. If p is a root of a2y2+by + c = 0, q is the root of a2y2+by + c = 0 and 0 <p< q, then the equation a2y2+by + c = 0 has a root r that always satisfies
(a) r = (p+q)/2
(b) r = p + (q/2)
(c) r = a
(d) p <r <q
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Step-by-step explanation:
Take f(x)=a
2
x
2
+2bx+2c
Now, f(α)=2(a
2
α
2
+bα+c)−a
2
α
2
……(1)
Since α is a root of a
2
x
2
+bx+c
So (1) becomes −a
2
α
2
Then f(β)=−2(a
2
β
2
−bβ−c)+4a
2
β
2
……(2) which similarly becomes f(β)=2a
2
β
2
Since f(α)f(β)=−4α
2
β
2
a
4
<0
Hence the root of f(x) ie γ lies between α and β
i.e. α<γ<β.
Step-by-step explanation:
[tex]p \leqslant r \leqslant q \\ [/tex]
Answer is option d