Can You Solve for Y Using the Quadratic Formula?

[1MileCrash]

x=-6y^2 + 4y

 

[mathman]

Use the quadratic formula.

 

[zhermes]

Try completing the square (i.e. as-if it was just a normal quadratic equation of a single variable)... it should work.

 

[1MileCrash]

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.

 

[gsal]

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