(x-1)^2 + (x+2)^3 >= 0
Use foil to simplify.
x^2 -2x +1 + x^3 + 6x^2 + 12x + 8 >=0
(For any a and b, (a+b)^2 = a^2 + 2ab + b^2 and (a+b)^3 = a^3 + 3ab^2 + 3a^2b + b^3)
Combine like terms.
x^3 + 7x^2 + 10x + 9 >=0
However...I cannot find any rational roots from it. Does your teacher allow you to use a calculator to find roots?
(x-1)^2 + (x+2)^3 >= 0
hellppp?! how do i solveee?
8 answers
no, but the directions above the question say to solve the inequality and write the solution in interval notation
no were not allowed to use calculators
Hmmmm. Can you graph it with a calc? You can see where it is positive.
I think it is positive for any x greater than x = -5.5 approximately. It has a minimum at x = -.88 but that above the x axis
A cubic has to have at least one real root. It starts negative and ends positive if the coefficient of x^3 is positive. If it has more than one, it has three. I believe this one has only one and the other two are complex.
Chrissy, could you verify if the second term is to the third power or is it just squared? They are completely different problems.
I will assume the second term is still cubed.
You probably mean you cannot use calculators to solve the equation directly? I hope you are allowed to use calculators to do algebraic calculations. If that's the case, you can find the root numerically.
Newton's method can be used if you have already done differential calculus. The following procedure assumes that you have not.
By putting f(x) = (x-1)^2+(x+2)^3, we evaluate
f(-5.5)=-0.625
f(-5.4)=1.656
By interpolation, we find new x=-5.4726
Interpolating between f(-5.4726) and f(-5.474), we find
x=5.473402 which compares favourably with the theoretical value of -5.4734020218...
A sketch of the graph of
f(x)=(x-1)^2 + (x+2)^3, as well as that of (x+2)^3 and (x-1)^2 is shown at the following link.
http://i263.photobucket.com/albums/ii157/mathmate/Chrissy.png
It will be noted that as indicated by Damon, there is a minimum at x=-1 above the x-axis, hence there is only one real root. Therefore we conclude:
f(x)≤0 for [-5.4734...,+∞)
I will assume the second term is still cubed.
You probably mean you cannot use calculators to solve the equation directly? I hope you are allowed to use calculators to do algebraic calculations. If that's the case, you can find the root numerically.
Newton's method can be used if you have already done differential calculus. The following procedure assumes that you have not.
By putting f(x) = (x-1)^2+(x+2)^3, we evaluate
f(-5.5)=-0.625
f(-5.4)=1.656
By interpolation, we find new x=-5.4726
Interpolating between f(-5.4726) and f(-5.474), we find
x=5.473402 which compares favourably with the theoretical value of -5.4734020218...
A sketch of the graph of
f(x)=(x-1)^2 + (x+2)^3, as well as that of (x+2)^3 and (x-1)^2 is shown at the following link.
http://i263.photobucket.com/albums/ii157/mathmate/Chrissy.png
It will be noted that as indicated by Damon, there is a minimum at x=-1 above the x-axis, hence there is only one real root. Therefore we conclude:
f(x)≤0 for [-5.4734...,+∞)
The last line should read:
f(x)≥0 for x∈ [-5.4734...,+∞)
f(x)≥0 for x∈ [-5.4734...,+∞)