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

Answer:

4

Step-by-step 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

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

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

$2p + 2± \frac{ \sqrt{8p - 28} }{2}$

$2p + 2± \frac{2 \sqrt{2p - 7} }{2}$

$2p + 2± \sqrt{2p - 7}$

So the roots are

$2p + 2 + \sqrt{2p - 7}$

and

$2p + 2 - \sqrt{2p - 7}$

We know the distance between these two points are

2 so

$|(2p + 2) + \sqrt{2p - 7} - (2p + 2) - \sqrt{2p - 7} | = 2$

$| \sqrt{2p - 7} + \sqrt{2p - 7} | = 2$

$|2 \sqrt{2p - 7} | = 2$

$2 \sqrt{2p - 7} = 2$

$p = 4$

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.