Question

let a and b be the roots of the equation x^2-2(p+1)x+p^2+8=0. If the absolute value of the difference of a and b equals to 2. find p

Answer 1

4

Explanation:

a and B are the roots of

`{x}^(2) - 2(p + 1)x + {p}^(2) + 8 = 0`

`{x}^(2) - 2px - 2x + {p}^(2) + 8 = 0`

This look like a parabola since p will be a constant so let use a quadratic formula

`{x}^(2) -(2p + 2)x + {p}^(2) + 8 = 0`

`2p + 2± \frac{ \sqrt{4 {p}^(2) + 8p + 4 - (4 {p}^(2) + 32) } }{2}`

`2p + 2± ( √(8p - 28) )/(2)`

`2p + 2± (2 √(2p - 7) )/(2)`

`2p + 2± √(2p - 7)`

So the roots are

`2p + 2 + √(2p - 7)`

and

`2p + 2 - √(2p - 7)`

We know the distance between these two points are

2 so

`|(2p + 2) + √(2p - 7) - (2p + 2) - √(2p - 7) | = 2`

`| √(2p - 7) + √(2p - 7) | = 2`

`|2 √(2p - 7) | = 2`

`2 √(2p - 7) = 2`

`p = 4`

Let plug it in to see.

`{x }^(2) - 2(4 + 1)x + {4}^(2) + 8 = 0`

`{x}^(2) - 10x + 24 = 0`

`(x - 6)(x - 4) = 0`

`x = 4`

`x = 6`

So they have a distance of 2.