how can i solve this???

[DPHJetta426]

how can i solve this?

4x^2 + 72x + 65 = 4y^2 + 4y + 1

the answer i need is x = 2 and y = 7. i haven't had a math course in years and i'm not even sure if it is possible to do without just plugging in numbers until i get it. any help would be appreciated.

[opus]

This equation does have more then 1 solution (x=0 and y=3,53 (roughly))!
But I also only tried individual values!

[anna69]

Put in any value you like for y. That makes the right hand side a value, let's call it capital-Y

Then you have 4x^2 + 72x + 65 = Y
or 4x^2 + 72x +(65-Y) =0 which is quadratic in x, solved in the normal way. That quadratic will have 2 real roots, or 1 real root, or no real roots depending on the effect of the Y part.

[Zachtk8702]

This is what I got when I tried to work it.............. If you know what X nad Y are just plug them In.



4x^2 + 72x + 65 = 4y^2 + 4y + 1


72x +4x^2 =88x
4^2y=16 + 4y = 20y

88X + 65 = 20y+1

Now to find 1 varible you will replace the other with zer0

88(0)+65 = 20y +1
65=20y+1
Subtract the 1 from both sides
64 = 20y
3.2 = Y

Now you have Y

88X + 65 = 20(3.2)+1
20(3.2) = 64

88X + 65 = 64
X = -81


I think thats right..... Its been a while May have to replace each with 1 instead of zero cant remember.
-65 -65
88x = -1

[azteched]

You have 2 variables but only one equation - if you want unique solutions, you need 2 equations.

[Ecniv]

Um dunno. Did quadratic equations around 10 years ago...

If the x side was 0 then y = -0.5
if y side is 0 then
x = (-7 + sqrroot(-991)) / 8
x = (-7 - sqrroot(-991)) / 8

No scientific calc with sqr root on it here... so not sure what miniscule amount that comes to.

This of course is probably completely wrong, and you should somehow get what x equals in terms of y. But that blew my brain so I'm giving up.

If you did it ages ago and are only coming across it now I would guess its because someone is doing said quadratic equations and they should have a book you can read... then post here with the answer

Pls post the answer!!

[anna69]

you should somehow get what x equals in terms of y.

Which is what I said ages ago.... stick any "y" value in, do the right hand side which now a constant (for any "y" value), now you have a quadratic in "x" solved in the normal way....

4x^2 + 72x +(65-Y) =0

The roots of a quadratic are x= (-b +- sqrt((b^2)-4ac)) / 2a

where in this case a=4, b=72 and c=(65-Y)
where Y was the result of the right hand side for any value of y

[NotLKH]

If you mean to find the sets of Integral (x,y) such that they solve your equation, then you could proceed in this manner:

Consider both sides of your equation "4x^2 + 72x + 65 = 4y^2 + 4y + 1"
in the form U^2 + k = V^2
Obviously, V = (2y+1)
Now, if we say U = aX+b, and we square it, we see that

a²x + 2abx + b² + k = 4x^2 + 72x + 65

So a² = 4, or:
a = 2

2ab = 72; ab = 36; 2b = 36, or:
b = 18

b² + k = 65;324 + k = 65; or:
k = -259

Therefore your equation "4x^2 + 72x + 65 = 4y^2 + 4y + 1", which we've remapped to U^2+k=V^2, becomes:

(2X+18)² - 259 = (2Y+1)²
or rearranged:

(2X+18)² - (2Y+1)² = 259

Which, since for any eq of the form M² - N² is identical to (M-N)*(M+N), then we now have

(2X+18-2Y-1)*(2X+18+2Y+1) = 259, or:
(2X-2Y+17)*(2X+2Y+19) = 259

See where I'm going?
Now, 259 is a composite multple of 7*37, both primes.
so either:

case 1) 2X-2Y+17=7 and 2X+2Y+19=37
or

case 2) 2X-2Y+17=37 and 2X+2Y+19=7

Both of which can be solved for their particular X, Y values.

Specifically, Case 1 produces X=2, Y=7
and Case 2 produces X=2, Y=-8














BTW, U in actuality could be = ⁺/_-(2X+18),
and similarly V could be = ⁺/_-(2Y+1),

So you have 4 combinations to solve from, not just the 1 that I illustrated.

[sql_lall]

(2X-2Y+17)*(2X+2Y+19) = 259

u said make it 7 * 37 or 37 * 1
BUT couldn't it also be 1 * 259, or 259 * 1?

e.g.
2x - 2y + 17 = 1
2x + 2y + 19 = 259
=> x = 56, y = 64

[NotLKH]

Good Point!

Sorry, I forgot to note that.