If 3 is a root of the quadratic equation x2−x+a=0, find the value of b so that the roots of the equation x2+a(2x+a+2)+b=0 are equal.
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Explanation:
Given, x2−(a−3)x+a=0
⇒D=(a−3)2−4a
=a2−10a+9
=(a−1)(a−9)
Case I:
Both the roots are greater than 2
D≥0,f(2)>0,−2ab>2
⇒(a−1)(a−9)≥0;4−(a−3)2+a>0;2a−3>2
⇒a∈(−∞,1]∪[9,∞);a<10;a>7
⇒a∈[9,10) ..........(1)
Case II:
One root is greater than 2 and the other root is less than or. equal to 2. Hence,
f(2)≤0
⇒4−(a−3)2+a≤0
⇒a≥10 ..........(2)
From (1) and (2)
a∈[9,10)∪[10,∞)⇒a∈[9,∞)
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