Can You Solve for Y Using the Quadratic Formula?

  • Context: High School 
  • Thread starter Thread starter 1MileCrash
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Discussion Overview

The discussion revolves around solving the equation x = -6y² + 4y for the variable y using the quadratic formula. Participants explore different methods, including the quadratic formula and completing the square, while considering the implications of their approaches on graphing the functions involved.

Discussion Character

  • Exploratory, Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant suggests using the quadratic formula to solve for y.
  • Another participant proposes completing the square as an alternative method.
  • A different participant contemplates the graphical representation of the functions, suggesting that plotting both equations will yield perpendicular graphs.
  • One participant reformulates the equation into standard quadratic form and applies the quadratic formula, deriving two expressions for y, while noting the restriction on x to avoid negative values under the square root.

Areas of Agreement / Disagreement

Participants express various methods for solving the equation, but there is no consensus on a single approach or resolution to the problem. Multiple competing views remain regarding the best method to use.

Contextual Notes

Participants note that the derived expressions for y depend on the condition that x cannot exceed 2/3 to maintain the validity of the square root. There may be additional assumptions or limitations regarding the domain of x that are not fully explored.

1MileCrash
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x=-6y^2 + 4y
 
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Use the quadratic formula.
 
Try completing the square (i.e. as-if it was just a normal quadratic equation of a single variable)... it should work.
 
okay, so plotting BOTH of these graphs will give me a perpendicular graph of y=-6x^2+4x, correct? each of the functions from the quadratic formula should be each half of a perpendicular parabola.
 
hhhmmm...I just took the equation and put it like this:

-6y2 + 4y - x = 0

and solved it using the quadratic formula, so

y = ( -b + sqrt(b2 - 4ac) ) / 2a

where

a=-6
b=4
c= -x

and the other solution to y is with the negative value of the square root.

I get something like this:

y = 1/3 + sqrt( (2-3x)/18 )

and, of course,

y = 1/3 - sqrt( (2-3x)/18 )

which tells me that x cannot be larger than 2/3, otherwise the radical becomes negative.

So, you can plot y versus x for -inf <= x <= 2/3
 

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