Roots of a quadratic equation

[Blake Andrews]

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.

 

[MarkFL]

If the roots are \(a\) and \(a+1\) then you can write:

[MATH]x^2-(k+2)x+(k+1)=(x-a)(x-(a+1))=x^2-(2a+1)x+(a^2+a)[/MATH]
Equating like coefficients, we obtain:

[MATH]k+2=2a+1\implies k=2a-1[/MATH]
[MATH]k+1=a^2+a\implies k=a^2+a-1[/MATH]
This implies:

[MATH]a^2+a-1=2a-1[/MATH]
You have this equation, but your next step should be:

[MATH]a(a-1)=0[/MATH]

 

[Blake Andrews]

Thank you, i redid that many times and made the same silly error each time.

 

[Steven G]

I think your logo is fine. However in x^2-(k+2)x+k+1=0, the b value is NOT (k+2) but rather it is ____?

 

[MarkFL]

Thank you, i redid that many times and made the same silly error each time.

I bet just about everyone here can relate to that.

 

[HallsofIvy]

Not me. I newer make silly errors!

 

[Steven G]

Not me. I newer make silly errors!

So you only make major errors?

 

[Blake Andrews]

I think your logo is fine. However in x^2-(k+2)x+k+1=0, the b value is NOT (k+2) but rather it is ____?

I had -b/a=k+2
So b is -(k+2)

 

[Steven G]

I think your logic is fine. However in x^2-(k+2)x+k+1=0, the b value is NOT (k+2) but rather it is ____?

I had -b/a=k+2
So b is -(k+2)

I am not sure what you mean when you wrote -b/a=(k+2) since that means that b = -a(k+2). I am not going to look back but possibly a=1??

You can simply read off b from x^2-(k+2)x+(k+1)=0. b=-(x+2). No need to even mention a.

 

[Blake Andrews]

I am not sure what you mean when you wrote -b/a=(k+2) since that means that b = -a(k+2). I am not going to look back but possibly a=1??

You can simply read off b from x^2-(k+2)x+(k+1)=0. b=-(x+2). No need to even mention a.

Sorry disregard that, yes b=-(k+2) straight from the equation.