need help with this problem

[Guest]

show that:
x^2+x+k is prime,
x^2+(k+1)x +k is reducible for all positive intergers k.

 

[stapel]

I don't know how rigourous your class is, but would this work?

If x² + x + k weren't prime, then it would factor over the reals, which means that, after setting it equal to zero, it would have real solutions.

So set it equal to zero, plug it into the Quadratic Formula, and show that you don't get any real solutions.

Eliz.

 

[soroban]

Hello, bryan-duong!

Show that: \(\displaystyle x^2\,+\,x\,+\,k\) is prime for all positive integers \(\displaystyle k.\)

Using the Quadratic Formula, <u>any</u> quadratic \(\displaystyle ax^2\,+\,bx\,+\,c\) can be factored.

. . \(\displaystyle ax^2\,+\,bx\,+\,c\;=\;a\left(x\,-\,\frac{-b\,+\,\sqrt{b^2 - 4ac}}{2a}\right)\left(x\,-\,\frac{-b\,-\,\sqrt{b^2 - 4ac}}{2a}\right)\)


The given quadratic has: \(\displaystyle a = 1,\,b = 1,\,c = k.\)
. . Its discriminant is: \(\displaystyle 1\,-\,4k\), which must be nonnegative.
. . Hence: .\(\displaystyle 1\,-\,4k\,\geq\,0\qquad\Rightarrow\qquad k\,\leq\,\frac{1}{4}\)

Since \(\displaystyle k\) is a positive integer, this is impossible.

Show that: \(\displaystyle x^2\,+\,(k+1)x\,+\,k\) is reducible for all positive intergers \(\displaystyle k.\)

The quadratic factors: .\(\displaystyle (x\,+\,1)(x\,+\,k)\)

 

[Guest]

thanks for your responses. They help me alot =)