c) The equation x2 + ax+b=0 has two real roots a and B. Show that Xx+1=-(axy +b)/xk is convergent near x = a if|a|>IBI.
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Answer
Correct option is
A
b=0,a>0
x
2
+ax+b=0 has distinct real roots.
Hence
a
2
−4b≥0 ...(i)
a
2
≥4b
aϵ(−∞,−2
b
]∪[2
b
,∞)
And
∣x∣
2
+a∣x∣+b=0 has only one real root.
Then
(∣x∣+
2
a
)
2
−
4
a
2
+
4
4b
=0
(∣x∣+
2
a
)
2
=
4
a
2
−4b
∣x∣=
2
a±
a
2
−4b
Now
a
2
−4b
<a (considering b as positive)
Yet we have only one real root.
Now a cannot be negative in any case, as negative value of a wont give any root in second case.
Hence b=0 and a>0
∣x∣ can only be positive,
Hence
x=
2
a+
a
2
−4b
x=
2
a+a
x=a
Step-by-step explanation:
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