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 eg x^4+3x^2-4 y=2 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)

WillDecember 21, 2016 at 11:20 amIs there a method to figure out what to let y be equal to because sometimes I find it hard to.

Jason ShackelDecember 21, 2016 at 1:19 pmall of them are y =, in all of the questions y is shown as the 0

lololJanuary 1, 2017 at 9:15 pmits not a method but its always the power of the second term

eg

x^4+3x^2-4

y=2

as in 3x^2

the power is two

johnFebruary 14, 2017 at 1:03 pm1/3^2 is = 1/9

how did you get 2/3 to be y^2

Stuart the ExamSolutions GuyFebruary 14, 2017 at 8:16 pmy^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)