for example, how would u solve, y cubed = 40....i want to know how to solve it step by step, not using a calculator to use cubic root....THANKS
-
y³ = 40
y = 40^(1/3)
y = 2∙5^(1/3)
The answer in exact form is 2 cube roots of 5. That's it. Calculators only give you a decimal approximation.
y = 40^(1/3)
y = 2∙5^(1/3)
The answer in exact form is 2 cube roots of 5. That's it. Calculators only give you a decimal approximation.
-
y^3=40
First you have to factor out a y from the equation.
y(y^2)=40
Then you will set each group equal to 40 and solve.
y=40,
y^2=40
One of your answers will be y=40, and for the other, you will take the square root of y and the square root of 40.
y= square root of 40, or y=(40)^1/2, and simplified is y=2(10)^1/2. But since you are taking the square root, your possible answers are positive and negative.
So your final answers will be y=40 and y=plus or minus square root of 40
First you have to factor out a y from the equation.
y(y^2)=40
Then you will set each group equal to 40 and solve.
y=40,
y^2=40
One of your answers will be y=40, and for the other, you will take the square root of y and the square root of 40.
y= square root of 40, or y=(40)^1/2, and simplified is y=2(10)^1/2. But since you are taking the square root, your possible answers are positive and negative.
So your final answers will be y=40 and y=plus or minus square root of 40
-
³√(40) = 2³√(5). You cannot do any better without taking numerical approximations (calculator, Newton's method, etc)
-
y^3 - 1 = 0
(y - 1) (y^2 + y + 1) = 0
y = 1
(y - 1) (y^2 + y + 1) = 0
y = 1
-
you should find a way to factor the eq. like:
ax³ + bx² + cx + d = 0
(x - p)(x - q)(x - r) = 0
where x1 = p ; x2 = q ; x3 = r
ax³ + bx² + cx + d = 0
(x - p)(x - q)(x - r) = 0
where x1 = p ; x2 = q ; x3 = r