ardz24 wrote:If 4 is one solution of the equation x² + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

Let's first determine the value of k.

Since x = 4 is a solution to the equation x² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is x² + 3x + (-18) = 10
This is the same as: x² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: x² + 3x - 28 = 0
Factor: (x + 7)(x - 4) = 0
So, x = -7 or x = 4

We already know that x = 4 is one solution.
So, the other solution is x = -7

Answer: A

Cheers,
Brent

BTGmoderatorAT wrote:If 4 is one solution of the equation x^2 + 3x + k = 10, where k is a constant, what is the other solution?

(A) -7
(B) -4
(C) -3
(D) 1
(E) 6

I'm confused how to set up the formulas here. Can any experts help?

Since x = 4, we have:

4^2 + (3)(4) + k = 10

16 + 12 + k = 10

28 + k = 10

k = -18

Next we plug -18 into the given equation for k and then solve for x.

x^2 + 3x - 18 = 10

x^2 + 3x - 28 = 0

(x + 7)(x - 4) = 0

x = -7 or x = 4

Thus, -7 is the other solution.

Answer A