What is the sum of the roots of the equation
(x^2-4x+4)(x+2)(x-8) = 0
I thought it was 8 but the teacher said it's wrong
Discriminant of equation x² - 4 x + 4 is ∆ = ( - 4 )² - 4 ∙ 1 ∙ 4 = 16 - 16 = 0
If the discriminant is equals zero, the roots are equal.
That is called a double root.
Root of x² - 4 x + 4 = 0 is 2 multiplicity two
Root of x + 2 = 0 is - 2
Root of x - 8 = 0 is 8
So the sum of the roots = 2 + 2 - 2 + 8 = 10
or, just factor the whole thing, the last part is already done.
(x^2-4x+4)(x+2)(x-8) = 0
(x-2)(x-2)(x+2)(x-8)
the roots are, 2, 2, -2, 8
that sums to 10
To find the sum of the roots of the given equation, you need to find the roots first. Let's break down the equation and solve it step by step.
The equation is: (x^2 - 4x + 4)(x + 2)(x - 8) = 0
1. Start by factoring the quadratic term, x^2 - 4x + 4. This quadratic can be factored as a perfect square: (x - 2)^2.
The equation becomes: (x - 2)^2(x + 2)(x - 8) = 0
2. Set each factor equal to zero and solve for x:
(x - 2)^2 = 0 (already factored)
x + 2 = 0 --> x = -2
x - 8 = 0 --> x = 8
So we have three roots: x = 2, x = -2, and x = 8.
Now, let's find the sum of these roots:
Sum of roots = 2 + (-2) + 8 = 8
You are correct. The sum of the roots is indeed 8. If your teacher gave a different answer, it is possible that there was a miscommunication or error in their explanation. You may want to clarify with your teacher to ensure that there is no misunderstanding.
Or, you can expand it out and use the fact that the sum of the roots of
x^4 + ax^3 + bx^2 + cx + d is -a
(x^2-4x+4)(x+2)(x-8) = x^4 - 10x^3 + 12x^2 + 40x - 64
so the sum of the roots is 10