# Square equation problem

• Sep 29th 2009, 01:52 PM
matt12345
Square equation problem
Hello
I have equation:
Code:

`x^2 - px + q = 0`
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).
• Sep 30th 2009, 09:49 AM
aidan
Quote:

Originally Posted by matt12345
Hello
I have equation:
Code:

`x^2 - px + q = 0`
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).

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```
That's eight solutions.
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, 08:14 AM
matt12345
Quote:

Originally Posted by matt12345
Hello
I have equation:
Code:

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

x1 and x2 are roots
Root (mathematics) - Wikipedia, the free encyclopedia

Sorry for my English
• Oct 16th 2009, 07:17 AM
matt12345
ref
• Oct 16th 2009, 08:24 AM
pankaj
Condition 1:

Since roots are real and distinct:

$Discriminant>0$

$p^2-4q>0$

$4q................................(1)

Condition 2:

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

$p<10$

Condition 3:
Let $f(x)=x^2-px+q$

$f(5)>0$

$25-5p+q>0$

From conditions 1,2 and 3 we can see that $0 and $0,where p and q are integers

I hope this helps.
• Oct 20th 2009, 07:56 AM
matt12345
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, 12:45 PM
pankaj
Since $x_{1}$ and $x_{2}$ are roots of $f(x)=0$,so $(x-x_{1})$ and $(x-x_{2})$ are factors of f(x)=0

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

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

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

Since $4q

$0

$\therefore 4q

due to which $0

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

$5p<25+q<25+25$

$p<10$.

Thus the values $0 and $0 also satisfy the condition $f(5)>0$
• Oct 21st 2009, 11:03 AM
matt12345
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, 07:03 PM
pankaj
I do not understand your querry