Power Prep Question - DC Quadratic

[bilon.geiger]

One of the roots of the equation x^2+KX-6=0 is 3, and K is a constant.

Quantity A: The value of K
Quantity B: -1

Please explain how to solve. I do not understand how to think through and solve this problem.

[tommywallach]

One of the roots of the equation x^2+Kx-6=0 is 3, and K is a constant.

Quantity A: The value of K
Quantity B: -1

Hey Bilon,

So when they tell you that one of the roots is 3, we already know one set of the two parentheses that we'll need in order to un-FOIL this quadratic:

(x-3) * ???? = x^2 + KX - 6

Now if you know how quadratics work, you also know that the two numbers in the parentheses will multiply to the constant (in this case, -6). Because we already have -3, we know the other number must be 2:

(x - 3) * (x + 2) = x^2 + Kx - 6

Now we can solve for K, which will simply be the sum of the two numbers in parentheses:

(x - 3) * (x + 2) = x ^ 2 - x - 6

K = -1.

The answer to the question is C.