If one root of the quadratic equation `px^2 +qx + r = 0 (p != 0)` is a surd `sqrta/(sqrta+sqrt(a-b),` where `p, q, r; a, b` are all rationals then the other root is -

Options

A. `(sqrt(a))/(sqrt(a)- sqrt(a)-b)`
B. `(sqrt(a)-sqrt(a)-b)/(sqrt(b))`
C. `a+ sqrt(a(a-b))/(b)`
D. `(a+ sqrt(a(a-b)))/(b)`

Answer

1,4

Explanation

One root `(sqrt(a))/(sqrt(a)+sqrt(a)-b)`
On retionalizing , we get
`(sqrt(a))/(sqrt(a)+sqrt(a)-b)xx(sqrt(a)-sqrt((a)-b))/(sqrt(a)+sqrt((a)-b))`
`(sqrt(a) (.sqrt(a) - sqrt(a-b)))/(b)= (a - sqrt(a (a - b)))/(b)`
` therefore ` Other root = ` (a+sqrt((a-b)))/(b)`
Rationalizing , we get
` (a+sqrt((a-b)))/(b)xx(a - sqrt(a(a-b)))/(a - sqrt(a(a-b)))`
`(a^(2) - (a^(2) - ab))/(b(a- sqrt(a (a - b))))= (sqrt(a))/(sqrt(a)- sqrt(a-b))`.