tuan1812

According to a question in Barron's book:
The sum of roots of the equation:
(x-a)^2(x-b)(x-c)
is: a + a + b + c.

Is it correct? I think the correct answer should be a+b+c. Please help!!

uvaismydream

Yeah that's correct imo.

(x-a)^2 = (x-a)(x-a) therefore:

(x-a)(x-a)(x-b)(x-c)
Roots: a, a, b, c
Sum of roots: a+a+b+c

stix2400

I would argue that it should be a+b+c

Just consider (x-2)^2=0

This is only true for x=2

So there may be a double root at x=2 but still, the number 2 is the only value which satisfies the equation. The answer of 4 would be dumb.

Just like the October SAT question this year. for how many values is (x+6)^2=0. Well, 1. Arguing a+a+b+c would be like arguing that there are two values of x which satisfy the equation, -6 and -6.

Barron's should have reworded the question. Collegeboard would ask that only to mean a+b+c, I promise.

tuan1812

Yeah, I think a+b+c also. Thanks a lot!
There are also some other minor mistakes in Barron's book, but I think this one is the biggest

Chapie

No there isn't any mistake!
Consider this:
(x-3)^2=0
You would say that the only root is 3 but it isn't
(x-3)^2=0 =>
(x-3)(x-3)=0
There are 2 roots not one!But both are 3

Chapie

" stix2400

Just like the October SAT question this year. for how many values is (x+6)^2=0. Well, 1. "


Well this is something different!
Lets take again this example:
(x-3)^2=0
There are two roots and both have value of two.So the only value that satisfies the equation is 2!
Hope to help you

stix2400

Yes, yes I'm well aware of the double root thing.

But I'm pretty sure the better answer is a+b+c and if collegeboard were to ask this question I would put that.

They'd probably just ask "what is the sum of the possible solutions of the polynomial" though, because that's a much better worded question.

SwaGGeReR

I've come across a problem like this before and put a+b+c, which was wrong. When you have something squared, there are TWO roots, albeit they're the same. Therefore, when a question asks for the sum of the roots, you must account for BOTH the roots of the square.

Sci-Fry

Check out Viete's theorem. That discusses the sum of roots and I think it follows that this would be a+a+b+c.

estrat1

I would calculate it as a + b + c, but, unfortunately, I'm convinced that that's not the way CollegeBoard would do it.

If we say f(x) = (x-a)^2, and the roots are all the values of x that satisfy f(x) = 0, we would have x belongs to {a} and therefore the sum of the roots should be a.

On the other hand, we could say that for any polynomial x^2 - sx + p = 0, the sum of the solutions should be x1 + x2 = s (I believe this is what Sci-Fry is referring to). Therefore, for a polynomial (x-a)^2 = 0, we have x^2 - 2ax + a^2 = 0 => the sum of the roots should be -(-2a) = 2a.

I kind of dislike the second way of doing things, but I also kind of understand it.