Consider a quadratic equation of the form ax^2+bx+c=0 with roots p and q.

Sum of roots = p+q
Product of roots = pq

a=1, b=-(k+1) and c=1

For p and q to be real and distinct,
b^2-4ac > 0

Solving for k gives k>1 or k<-3

Using the well known quadratic formula x = [ -b ± sqrt(b^2-4ac)] / 2a with p and q as the roots then :

p = [ -b + sqrt(b^2-4ac)] / 2a and q = [ -b - sqrt(b^2-4ac)] / 2a

It can be easily shown that p+q = -b/a. and pq = c/a (1)

For p and q to be real and distinct, the discriminant b^2-4ac > 0

then (b^2-4ac)/a^2 > 0 since a^2 > 0 the inequalities still hold
(b/a)^2 - 4(c/a) > 0

Given that p+q = k +1 and pq=1 (2)

Substituting from (1) and (2) above gives (k+1)^2 - 4 > 0
Solving for k gives k > 1 or k < -3
Hope this help!

thank you!

Hi Kristen,
I have an alternative solution which you will find easier to understand.
Using the algebraic identity :
(p+q)^2 = p^2 + q^2 + 2pq

And by subtracting 4pq from both sides gives :
(p+q)^2 - 4pq = p^2 + q^2 +2pq - 4pq = (p-q)^2

Since p and q are real and distinct, (p-q)^2 is strictly positive.

Hence (p+q)^2 - 4pq > 0

Substituting p+q=k+1 and pq=1 into the above inequality gives :
(k+1)^2 - 4(1) > 0

Solving for k gives k > 1 or k < -3
Giving the same answer as previous method.