Is this how you solve t(t+1)(t+2)=0, x^3+4x=4x^2, and 3r^2=4r-1?
t(t+1)(t+2)=0
t=0 or t= -1 or t=2
{0, -1, -2}
x^3+4x=4x^2
x^3+4x-4x^2=0
x(x^2+4-4x)=0
x(x^2-4x+4)=0
x(x-2)(x-2)=0
x=0 or x=2
{0,2(double root)}
3r^2=4r-1
3r^2-4r+1=0
r^2-4r+3=0
(r-3)(r-1)=0
r=3 or r=1
{1,3}
The first two are fine.
The third one: 3r^2 - 4 + 1 = (3r-1)(r-1)
Take it from there.
thanks
it was {1/3, 1} right
Yes, you have correctly solved the given equations. To solve these equations, you have used various algebraic manipulation techniques such as factoring and setting equations equal to zero.
For the equation t(t+1)(t+2)=0, you have factored it into three terms and set each term equal to zero. This allows you to solve for the values of t that make the equation true, which are t=0, t=-1, and t=2. Therefore, the solution set is {0, -1, 2}.
For the equation x^3+4x=4x^2, you rearranged the equation by subtracting 4x^2 from both sides to get x^3+4x-4x^2=0. Then, you factored out an x term to obtain x(x^2+4-4x)=0. Further simplifying, you factored the quadratic expression x^2+4-4x into (x-2)(x-2) using factoring techniques. This gives you x(x-2)(x-2)=0. By setting each factor equal to zero, you find the values x=0 and x=2. However, x=2 is a double root since it appeared twice in the factored form. Therefore, the solution set is {0, 2}.
Lastly, for the equation 3r^2=4r-1, you rearranged the equation by subtracting 4r and adding 1 to both sides to get 3r^2-4r+1=0. You then proceeded to factor the quadratic expression r^2-4r+3 into (r-3)(r-1). By setting each factor equal to zero, you find the values r=1 and r=3. Therefore, the solution set is {1, 3}.