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
Got the answer:)
To find the roots of the equation x^3 + 8x^2 = 20x, you need to rearrange the equation and solve for x. Here's how you can do it:
1. Start by subtracting 20x from both sides of the equation:
x^3 + 8x^2 - 20x = 0
2. Now, factor out an x from each term:
x(x^2 + 8x - 20) = 0
3. To further factor the quadratic equation inside the parentheses, you need to find two numbers whose product is -20 and whose sum is 8. In this case, the numbers are 10 and -2.
x(x + 10)(x - 2) = 0
4. At this point, you have three possible roots:
x = 0 (from the x term)
x + 10 = 0 (from the first factor)
x - 2 = 0 (from the last factor)
5. Solve each of the equations:
For x + 10 = 0, you will subtract 10 from both sides to get x = -10.
For x - 2 = 0, you will add 2 to both sides to get x = 2.
So, the three roots of the equation x^3 + 8x^2 = 20x are x = 0, x = -10, and x = 2.
Now, moving on to the equation 2x^2 - 10x + 12 = 0, you have already done the factoring correctly. Here's a step-by-step breakdown:
1. Start with the factored form: 2(x - 3)(x - 2) = 0
2. Set each factor equal to zero:
x - 3 = 0, which gives x = 3
x - 2 = 0, which gives x = 2
So, the roots of the equation 2x^2 - 10x + 12 = 0 are x = 3 and x = 2. Your solution is correct!