If `a lt b lt c lt d`, then for any real non-zero `lambda`, the quadratic equation `(x-a)(x-c)+lambda(x-b)(x-d)=0`,has

Explanation

`alt blt clt d`
`mu(x - a) (x - c) + lambda (x - b) (x-d) = 0 `
Let the corresponding expression be
`f(x) = mu(x - a) (x - c) + lambda(x - b) (x - d)`
`f(a) = lambda (a-b) (a -d)`
`f(c) = lambda (c - b) (c - d)`
`rArr f(a) f(c) = lambda ^(2) (a - b) (a - d) (c - b) (c - d) lt 0 `
`rArr f(a) and f(c)` are of opposite signs
Hence, root of the equation lies between a and c . Therefore , the
roots are real for all real `mu and lambda`.