Question

if roots of x^2-(a-3)x+a=0 are such that both of them is greater than 2 then​

Answer 1

Answer:

ANSWER

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,−

2a

b

>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,∞)