practise test 5, question 12

[irs031]

The two roots to the equation x2 + ax + b = 0 sum to 1 and differ by 5. What is the value of a + b?

A) -7
B) -6
C) 1
D) 5
E) 6

Explanation:
Let x1 and x2 denote the two solutions. We are told that the following equations hold true:

x1 + x2 = 1
x1 – x2 = 5

We can eliminate x2 by adding the two equations. This gives 2x1 = 6 or x1 = 3. We can then solve for x2 from either of the two equations (the result will be the same). For instance, from the first equation, we get x2 = 1 – x1 = 1 – 3 = –2. These two solutions, namely 3 and –2, would appear in the factored form of the quadratic as (x – 3)(x – (–2)) = 0, or (x – 3)(x + 2) = 0. Multiplying out yields x2 – x – 6 = 0, so that a + b = (–1) + (–6) = –7.
The correct answer is A.

I don't quite understand the underlined portion of the explanation. x^2+ax+b = 0 is a quadratic equation and we find that the two solutions are x1=3 and x2= -2; So far so good. However if all signs in the quadratic x^2+ax+b are "+", then the equation should be (a+b)^2. Why then we have minuses in the factored form (x-3)(x-(-2)) ? If the factors are (x-3)(x-(-2)) shouldn't the original equation be x^2 - ax + b = 0 like in (a-b)^2?

I am kind of lost here and this portion of the problem prevented me at arriving at a correct solution.

[tommywallach]

Hey Irs,

Root just means solution. So if we know the solutions are 3 and -2, that would factor to:

(x - 3)(x + 2) = 0

Because in this equation, both x = 3 and x = -2 correctly equal to zero.

As for the + thing, you're thinking of a TOTALLY separate issue. That's one of the three quadratic identities:

(a+b)^2 = a^2 + 2ab + b^2

That only applies when there are two variables. There's only one variable in this question.

-t

[irs031]

Tommy,

Thanks. What I believe mainly threw me off on this problem were the signs of the original equation. Special products such as (x+y)^2 do not apply here, that's true. However the original equation is given as x^2+ax+b = 0 ; When I saw a quadratic equation with all "+" sign I expected all factors to be positive as well. For example x^2+14x+33=0 has factors (x+11)(x+3). The factors instead turned out to be not all positive (x-3)(x+2) and when foiled the resulting quadratic equation was x^2-x-6=0 (second and third terms have a negative sign, contrary to the original equation x^2+ax+b=0). I guess for that reason the problem is classified as "devilish".

[tommywallach]

Hey Irs,

Aha. I can see how that would confuse. Unfortunately, the + sign doesn't inherently mean a has to be positive. You just might be adding a positive number. (And in the question's defense, quadratics are usually presented in that format, with that + sign, even if the proceeding equation ends up with a negative coefficient on the x.)

Devilish indeed!

-t