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
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
3 answers
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
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
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
(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