Solving for a < 100

[silva]

(See Attached Image)

If a and b, and x are natural numbers, find the possible values of a if a < 100

How would I solve this? Please help :)

http://www.vbforums.com/attachment.php?s=&postid=1364367

[plenderj]

Well you'd check for which values the square root would give even whole number results.

[bugzpodder]

well just say any value of a works.

[silva]

1 < a < 99

A can be from 2 to 98 - I think.

How can I prove / reason this?

s.

[kedaman]

thats the formula to solve quadratic polynominals
x^2+bx-a=0
so in other words
a=x^2+bx
put for instance x=1 and for any b in [1,98] you get a in [2,99]

[opus]

But 1,98 doesn't sound likea natuaral number!

[kedaman]

we use commas as decimal delimiter here in Finland, but in math i think its better to stay to . because , is used in other contexts, like [a,b] which means from and including a to and including b

[opus]

Sorry

[silva]

OK, I now think 9 < a < 100


For a = 99 To 1 Step -1
For b = 11000 To 1 Step -1
x = ((b + Sqr((b ^ 2) + (4 * a))) / 2)
If InStr(1, x, ".") = 0 Then List1.AddItem "a: " & a & " b: " & b & " x: " & x
Next
Next


How could I explain this or is there a fault in my reasoning?

s.

[prog_tom]

Originally posted by kedaman
thats the formula to solve quadratic polynominals
x^2+bx-a=0
so in other words
a=x^2+bx
put for instance x=1 and for any b in [1,98] you get a in [2,99]

WRONG, that's not the equation for solving quadratics.

x = -b +- sqrt(b^2 + 4ac)/2a

is the Equation for solving quadratics.

[kedaman]

with polynomals ax^2+bx+c=0 yes, but we have the same roots for all n for nx^2+nbx +nc=0.

[silva]

Originally posted by prog_tom
WRONG, that's not the equation for solving quadratics.

x = -b +- sqrt(b^2 + 4ac)/2a

is the Equation for solving quadratics.

Yes, that's the quadratic formula, but there are other ways in which you can solve quadratics.

Anyway, I proved it - don't worry :p

Thanks for the help.

s.