<p>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.</p>
<p>Is it correct? I think the correct answer should be a+b+c. Please help!!</p>
<p>Yeah that’s correct imo.</p>
<p>(x-a)^2 = (x-a)(x-a) therefore:</p>
<pre><code> (x-a)(x-a)(x-b)(x-c)
</code></pre>
<p>Roots: a, a, b, c
Sum of roots: a+a+b+c</p>
<p>I would argue that it should be a+b+c</p>
<p>Just consider (x-2)^2=0</p>
<p>This is only true for x=2</p>
<p>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.</p>
<p>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.</p>
<p>Barron’s should have reworded the question. Collegeboard would ask that only to mean a+b+c, I promise.</p>
<p>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</p>
<p>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</p>
<p>" stix2400</p>
<p>Just like the October SAT question this year. for how many values is (x+6)^2=0. Well, 1. "</p>
<p>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</p>
<p>Yes, yes I’m well aware of the double root thing. </p>
<p>But I’m pretty sure the better answer is a+b+c and if collegeboard were to ask this question I would put that.</p>
<p>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.</p>
<p>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.</p>
<p>Check out Viete’s theorem. That discusses the sum of roots and I think it follows that this would be a+a+b+c.</p>
<p>I would calculate it as a + b + c, but, unfortunately, I’m convinced that that’s not the way CollegeBoard would do it.</p>
<p>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.</p>
<p>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.</p>
<p>I kind of dislike the second way of doing things, but I also kind of understand it.</p>