x^3 + 8x^2 = 20x
How do I find the root of ^ that equation?
2x^2-10x+12 = 0
^ I found that one by...
2(x-3)(x-2)
x-3 = 0
x = 3
x-2 = 0
x =2
^ Is that the correct way to solve for that one?
Your solution to the second equation is correct, and the procedure also.
For the first equation, think of it as:
x^3 + 8x^2 = 20x
x^3 + 8x^2 - 20x = 0
x(x^2 + 8x - 20) = 0
then x=0 is a solution, and you can get two more solutions from the part in parentheses in a similar way to the second equation.
Thanks so much MathMate =)
You're welcome, and keep up the good work!
To find the roots of the equation x^3 + 8x^2 = 20x, you can follow these steps:
Step 1: Move all terms to one side of the equation to set it equal to zero:
x^3 + 8x^2 - 20x = 0
Step 2: Factor out the common factor, x:
x(x^2 + 8x - 20) = 0
Step 3: Now, you need to factor the quadratic expression (x^2 + 8x - 20). To factor it, you need to find two numbers whose sum is 8 and whose product is -20. In this case, the numbers are 10 and -2:
(x + 10)(x - 2) = 0
Step 4: Set each factor equal to zero and solve for x:
x + 10 = 0 --> x = -10
x - 2 = 0 --> x = 2
So, the roots of the equation are x = -10 and x = 2.
Now, let's move on to the second equation, 2x^2 - 10x + 12 = 0:
Step 1: Factor the quadratic expression:
2(x^2 - 5x + 6) = 0
Step 2: Factor the quadratic expression inside the parentheses, which is (x^2 - 5x + 6). You need to find two numbers whose sum is -5 and whose product is 6. In this case, the numbers are -3 and -2:
2(x - 3)(x - 2) = 0
Step 3: Set each factor equal to zero and solve for x:
x - 3 = 0 --> x = 3
x - 2 = 0 --> x = 2
So, the roots of the equation are x = 3 and x = 2.
In both cases, you have correctly solved for the roots of the equations. Well done!