Solving Roots - Algebra

[cricketfan10]

Could anyone help with this question?

My thoughts were to set it up as simulatenous equations (in terms of x1 and x2), however my answer is still in terms of alegbra!

 

[MarkFL]

I would look at the fact that given a quadratic:

[MATH]f(x)=ax^2+bx+c[/MATH]
which has the roots \(x_1\) and \(x_2\) that:

[MATH]x_1x_2=\frac{c}{a}[/MATH]
[MATH]x_1+x_2=-\frac{b}{a}[/MATH]
What do you get when you apply this to the given quadratic?

 

[cricketfan10]

I would look at the fact that given a quadratic:

[MATH]f(x)=ax^2+bx+c[/MATH]
which has the roots \(x_1\) and \(x_2\) that:

[MATH]x_1x_2=\frac{c}{a}[/MATH]
[MATH]x_1+x_2=-\frac{b}{a}[/MATH]
What do you get when you apply this to the given quadratic?

Thanks - so using this:

I'm left with (P/(1998 - P)) , does that seem as far as I can get? Does the fact they're positive integer roots indicate we can reduce this further?

 

[MarkFL]

On second thought, let's observe that 1997 is a prime number, so given the roots are integers we can write:

[MATH]x^2+px+1997=(x+1)(x+1997)=0[/MATH]
or

[MATH]x^2+px+1997=(x-1)(x-1997)=0[/MATH]
What are the possible values for \(p\) and the roots associated with those values?

 

[cricketfan10]

On second thought, let's observe that 1997 is a prime number, so given the roots are integers we can write:

[MATH]x^2+px+1997=(x+1)(x+1997)=0[/MATH]
or

[MATH]x^2+px+1997=(x-1)(x-1997)=0[/MATH]
What are the possible values for \(p\) and the roots associated with those values?

Amazing! Thank you so much - Never seen a question that relied on the knowledge of this technique, with prime number as the 'c' term. That makes a lot of sense.