Question

If roots of the equation x 2 −(a−3)x+a=0 are such that at least one of them is greater than 2, then

Answer 1

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Given, x

2 −(a−3)x+a=0

⇒D=(a−3) 2 −4a

=a 2−10a+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; 2

a−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,∞)

hope it helps u...!!

Answer 2

SOLUTION:-

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