Find the range of `f(x)=sqrt(x-1)+sqrt(5-xdot)`

Answer

`x^(2) - 4abx - (a^(2) - b^(2))^(2) = 0`

Explanation

` (alpha + beta)^(2) = (alpha + beta)^(2)`
and `(alpha - beta)^(2) = (alpha + beta) ^(2) - 4alpha beta `
` (a + b )^(2) - 4 ((a^(2) + b^(2))/(2))`
` 2 ab - (a^(2) + b^(2))`
2 ab - (a^(2) + b^(2))`
= - (a - b)^(2)`
Now , the required equation whose roots are ` (alpha + beta)^(2) and (alpha - beta )^(2)` is
`x^(2) - {(alpha + beta)^(2) + (alpha + beta)^(2)} x + (alpha + beta)^(2) (alpha - beta)^(2) = 0`
or `x^(2) - {(a +b)^(2) - (a- b)^(2)} x - (a + b)^(2) (a-b)^(2) = 0`
or ` x^(2) - 4 abx - (a^(2) - b^(2))^(2) = 0` .