Hello

I have equation:Equation has 2 solutions. p and q are integer and positive. How many different p and q exsist if both solutions are less then 5 (x1,x2<5).Code:`x^2 - px + q = 0`

Sorry for my English. Please help. Cheers.

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- Sep 29th 2009, 12:52 PMmatt12345Square equation problem
Hello

I have equation:Code:`x^2 - px + q = 0`

Sorry for my English. Please help. Cheers. - Sep 30th 2009, 08:49 AMaidan
I do not understand this (x1,x2<5).

Could you explain what that means in this situation?

Since p & q are integers, x must be greater than zero, and an integer.

Code:

x p q

1 1 0

1 2 1

1 3 2

1 4 3

2 2 0

2 3 2

2 4 4

3 4 3

If you removed x=1, p=1, q=0 and x=2, p=2, q=0, you would still have 6 solutions.

Could you explain the restrictions so that only 2 work? - Oct 1st 2009, 07:14 AMmatt12345
x1 and x2 are roots

Root (mathematics) - Wikipedia, the free encyclopedia

Sorry for my English - Oct 16th 2009, 06:17 AMmatt12345
ref

please help - Oct 16th 2009, 07:24 AMpankaj
Condition 1:

Since roots are real and distinct:

$\displaystyle Discriminant>0$

$\displaystyle p^2-4q>0$

$\displaystyle 4q<p^2$................................(1)

Condition 2:

Since both roots are less than $\displaystyle 5$,the sum of the roots will be less than $\displaystyle 10$.

$\displaystyle p<10$

Condition 3:

Let $\displaystyle f(x)=x^2-px+q$

$\displaystyle f(5)>0$

$\displaystyle 25-5p+q>0$

From conditions 1,2 and 3 we can see that $\displaystyle 0<p<10$ and $\displaystyle 0<q<25$,where p and q are integers

I hope this helps. - Oct 20th 2009, 06:56 AMmatt12345
THX

Why http://www.mathhelpforum.com/math-he...5ff0f15d-1.gif and what does this give? Why q<25 ?

What will be the resualt? - Oct 20th 2009, 11:45 AMpankaj
Since $\displaystyle x_{1}$ and $\displaystyle x_{2}$ are roots of $\displaystyle f(x)=0$,so $\displaystyle (x-x_{1}) $ and $\displaystyle (x-x_{2}) $ are factors of f(x)=0

$\displaystyle f(x)=x^2-px+q=(x-x_{1})(x-x_{2})$

Since both the roots are less than $\displaystyle 5$,clearly both $\displaystyle (5-x_{1})$ and $\displaystyle (5-x_{2}) $ are positive.Thus

$\displaystyle f(5)=5^2-p(5)+q=(5-x_{1})(5-x_{2})>0$

Since $\displaystyle 4q<p^2$

$\displaystyle 0<p<10$

$\displaystyle \therefore 4q<p^2<100$

due to which $\displaystyle 0<q<25$

Now these values of $\displaystyle p$ and $\displaystyle q$ must also satisfy the condition $\displaystyle f(5)>0$

$\displaystyle 5p<25+q<25+25$

$\displaystyle p<10$.

Thus the values $\displaystyle 0<p<10$ and $\displaystyle 0<q<25$ also satisfy the condition $\displaystyle f(5)>0$ - Oct 21st 2009, 10:03 AMmatt12345
Thx

But I don't know how many p&q satisfy the condition x^2 - px + q = 0 ?

For exaple in this reason result is 16 (I have written program in c++ and check a lot of p & q, if condition was satisfied r++ and r was a resault).

How can I get that? - Oct 21st 2009, 06:03 PMpankaj
I do not understand your querry