Question 1196803
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kx^2-2kx+5 = 0 is in the form ax^2+bx+c = 0 with
a = k
b = -2k
c = 5


According to Vieta's Formulas, we know that for quadratics the sum of the roots is -b/a and the product is c/a


sum of roots = -b/a
product of roots = c/a


The value of -b/a in this case is:
-b/a = -(-2k)/k
-b/a = 2
So the sum of the roots is 2 and not 4. 
Your teacher made a typo somewhere.


For instance, if k = 7, then we have {{{kx^2-2kx+5 = 0}}} turn into {{{7x^2-14x+5 = 0}}} 
Turn to the quadratic formula to find the two roots being {{{(7+sqrt(14))/7}}} and {{{(7-sqrt(14))/7}}}
The sum of these two roots is 2.


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Here's a proof of Vieta's Formulas when applying the quadratic case.


Let p and q be the two roots of a quadratic
This means x = p and x = q are solutions to ax^2+bx+c = 0 aka x^2 + (b/a)x + (c/a) = 0
We can get everything to one side to get x-p = 0 and x-q = 0
Then apply the zero product property getting (x-p)(x-q) = 0


Then expand and follow these steps
(x-p)(x-q) = 0
x(x-q) - p(x-q) = 0
x^2-qx - px + pq = 0
x^2 - (p+q)x + pq = 0


Comparing that to x^2 + (b/a)x + (c/a) = 0 shows:
b/a = x coefficient = -(p+q) = sum of the roots
c/a = constant = pq = product of the roots


Then as one last final step we turn the 
b/a = -(p+q)
into
p+q = -b/a
to help match the original statement made about Vieta's formulas
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