Solve for y in 16y2 -24xy -300y +x2 + 400x = 0

  • Thread starter Thread starter vanmaiden
  • Start date Start date
AI Thread Summary
The equation 16y² - 24xy - 300y + x² + 400x = 0 is quadratic in y, allowing for the use of the Quadratic Formula to solve for y. The equation can be rearranged into the standard form ay² + by + c = 0, where a = 16, b = -24x - 300, and c = x² + 400x. Users in the discussion emphasize the need to isolate y correctly and acknowledge that there will be two solutions for y, both dependent on x. Some participants express confusion over the results obtained from the quadratic formula, indicating a need for further practice. The conversation highlights the importance of recognizing the quadratic nature of the equation for solving it effectively.
vanmaiden
Messages
101
Reaction score
1
solving for "y"

Homework Statement


Solve for y in 16y2 -24xy -300y +x2 + 400x = 0


Homework Equations


N/A


The Attempt at a Solution


16y2 -24xy -300y +x2 + 400x = 0
16y2 -24xy -300y = -x2 - 400x
y (16y - 24x - 300) = -x2 - 400x

When I get here, I can isolate the y but not the y
 
Physics news on Phys.org


vanmaiden said:

Homework Statement


Solve for y in 16y2 -24xy -300y +x2 + 400x = 0


Homework Equations


N/A


The Attempt at a Solution


16y2 -24xy -300y +x2 + 400x = 0
16y2 -24xy -300y = -x2 - 400x
y (16y - 24x - 300) = -x2 - 400x

When I get here, I can isolate the y but not the y
What you are calling y and y are the same thing, so you have not isolated y.

The equation is quadratic in y, so you can solve for y using the Quadratic Formula.
16y2 -24xy -300y +x2 + 400x = 0
<==> 16y2 + (-24x - 300)y + (x2 + 400x) = 0

This is now in the form ay2 + by + c = 0. Can you take it from here?
 


What mark44 is saying is that your equation is a quadratic in y, and there is a forumla for those.
 


hunt_mat said:
What mark44 is saying is that your equation is a quadratic in y

Mark44 said:
The equation is quadratic in y

lol :wink:
 


Perhaps I should have read mark44's post a little more closely...
 


Thank you all for the help

Mark44 said:
Can you take it from here?

I see what I should do now, but I continue to get weird answers when I plug it into the quadratic formula. I'm sure it's just too early for me to be doing this stuff lol. I'll continue trying.

Thank you
 


Not sure what you mean by "weird answers," but there will be two solutions for y, both involving x.
 
Back
Top