In this video I show you how to solve quadratic equations in some function of x.
Solve the following:
Is there a method to figure out what to let y be equal to because sometimes I find it hard to.
all of them are y =, in all of the questions y is shown as the 0
its not a method but its always the power of the second term
as in 3x^2
the power is two
1/3^2 is = 1/9
how did you get 2/3 to be y^2
y^2 is not 2/3. y^2 is [x^(1/3)]^2 which is [x^(1/3)][x^(1/3)] and then using the rule for indices you add the powers so
y^2 = x^[(1/3)+(1/3)]
which is x^(2/3)