Blake Andrews
New member
- Joined
- Feb 12, 2020
- Messages
- 11
Find the value(s) of k if the quadratic equation x2-(k+2)x+k+1=0 has consecutive roots.
Let the roots be a and a+1.
a+(a+1)=k+2
k=2a-1
a(a+1)=k+1
k=a2+a-1
a2+a-1=2a-1
a2-a-2=0
(a+1)(a-2)=0
a=-1, 2
k=2a-1
=+/-3
My textbook answer is +/-1 and i can't seem to see where i've gone wrong.
Let the roots be a and a+1.
a+(a+1)=k+2
k=2a-1
a(a+1)=k+1
k=a2+a-1
a2+a-1=2a-1
a2-a-2=0
(a+1)(a-2)=0
a=-1, 2
k=2a-1
=+/-3
My textbook answer is +/-1 and i can't seem to see where i've gone wrong.