Let `a,b in N, a != b` and the two quadratic equations `(a-1)x^2-(a^2+2)x+a^2+2a=0 and (b-1)x^2-(b^2+2)x+(b^2+2b)=0` have a common root. The value of `ab` is

Options

A. `4`
B. `6`
C. `8`
D. `10

Answer

C

Explanation

`(c )` `a gt 1`, `b gt 1`, `a ne b`
Let `alpha `is the common root
`(a-1)alpha^(2)-(a^(2)+1)alpha+a^(2)+2a=0` ………`(i)`
`(b-1)alpha^(2)-(b^(2)+2)alpha+(b^(2)+2b)=0` ………..`(ii)`
`(i)xx(b-1)-(ii)xx(a-1)`, we get
`(a-b)(ab-a-b-2)(alpha-1)=0`..........`(iii)`
If `alpha=1`, then `a=b=1`
So, `a ne 1`
From `(iii)`, we get
`ab=a+b+2`
Let `a gt b gt 1`
`:. b=1+(b)/(a)+(2)/(a) lt 3 implies b=2`, `a=4`
`:. ab=8`