SOLUTION: complete the square 25x^2-9y^2+50x+250=0 and show work. do I use 0 for the y?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: complete the square 25x^2-9y^2+50x+250=0 and show work. do I use 0 for the y?      Log On


   



Question 1040317: complete the square 25x^2-9y^2+50x+250=0 and show work. do I use 0 for the y?
Found 2 solutions by Aldorozos, ikleyn:
Answer by Aldorozos(172) About Me  (Show Source):
You can put this solution on YOUR website!
The quadratic equation has a format of ax^2+bx+c = 0
For the equation of 25x^2-9y^2+50x+250=0 to look like ax^2+bx+c = 0 is to consider -9y^2 as a constant just like 250 which is a constant. If we add these two constants we get 250-9y^2 for the simplicity we can look at 250-9y^2 as another constant like C. Then in this case we can rewrite the equation
25x^2-9y^2+50x+250 = 25x^2+50x +c
Now we use quadratic equation to solve this. x= [-50+and - sqrt(-50)^2 -4(25)(c)]/2(25)

x = [-50+&-sqrt2500-4(25)(250-9y^2)]/50 We can multiply this a little more. However, we won't be able to have any number for x unless we know what y is.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
complete the square 25x^2-9y^2+50x+250=0 and show work. do I use 0 for the y?
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25x%5E2-9y%5E2%2B50x%2B250 = 0.

Complete the square in the left side:

  25x%5E2-9y%5E2%2B50x%2B250  = 

= %2825x%5E2%2B50x%2B250%29 - 9y2 = 

= %285x%2B5%29%5E2+%2B+250+-+25 - %283y%29%5E2 = 

= %285x%2B5%29%5E2 - %283y%29%5E2 + 225.

Done. Completed.