Solve algebraically
cuberoot(t=12)-t=o
I don't understand your equation.
I'm knew to this and I don't know how to lay it out, but t + 12 is under a square root symbol then - t = 10 and I'm suppose to solve algebraically
To solve the equation algebraically, we need to isolate the variable "t" on one side of the equation.
First, let's rewrite the equation in a more standard form:
∛(t) - t = 12
To eliminate the cube root, we can raise both sides of the equation to the power of 3:
(∛(t) - t)^3 = 12^3
Expanding the left side using the binomial formula, we get:
(t) - 3(t)^2∛(t) + 3(t)^3 - t^3 = 1728
Combine like terms:
3(t)^3 - 3(t)^2∛(t) - t^3 + t = 1728
Now, let's simplify the equation further by letting u = ∛(t). Substituting, we have:
3u^3 - 3u^2u - u^3 + u = 1728
Simplify:
3u^3 - 3u^3 + u - u^3u = 1728
-u^2u -2u + u = 1728
-u^2u + 2u = 1728
Factor out u:
u(-u^2 + 2) = 1728
Now we have a quadratic equation:
-u^2 + 2 = 1728 / u
-u^2 + 2 = 1728u^(-1)
Moving all terms to one side:
u^2 - 2 + 1728u^(-1) = 0
Multiply through by u to get rid of the denominator:
u^3 - 2u + 1728 = 0
This is a cubic equation in terms of u. To solve it, you can use numerical methods or approximation techniques such as the Newton-Raphson method or graphing the equation to find the approximate solutions.