BTGmoderatorDC 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

There is no need to find the constant k ...

\[?\,\,\, = \,\,\,r\,\,\,\left( {{\text{other}}\,\,{\text{root}}} \right)\]
\[1{x^2} + 3x + \left( {k - 10} \right) = 0\,\,\,\,\,\,\,\,\left( {k\,\,{\text{cte}}} \right)\]
\[4 + r = {\text{Sum}}\,\, = - \frac{3}{1}\,\,\,\, \Rightarrow \,\,\,r = - 7\]

Reminder:
\[a{x^2} + bx + c = 0\,\,\,\,\,\left( {a \ne 0} \right)\]
\[{\text{Sum}}\,\,{\text{of}}\,\,{\text{the}}\,\,{\text{roots}}\,\, = \,\, - \frac{b}{a}\,\,\,\,\,\,\,\,\left( {\Delta \geqslant {\text{0}}} \right)\]

The above follows the notations and rationale taught in the GMATH method.


POST-MORTEM: after reading the 2 solutions posted after mine, I would like to explain my solution a little better, so that there is no doubt (for all readers) that it is perfect. (*)

1. Given the equation \[1{x^2} + 3x + \left( {k - 10} \right) = 0\,\,\,\,\,\,\,\,\left( {k\,\,{\text{cte}}} \right)\] there is no question about the 3 "fundamental" coefficients involved:
\[a = 1\,\,\,,\,\,\,b = 3\,\,\,,\,\,\,c = k - 10\,\,\,\,\,\,\,\,\,\left( {k\,\,{\text{cte}}} \right)\]
2. A careful reading of the question stem permits us to conclude that this equation has two real roots. (We do not expect the other to be 4, and 4 by the way is not among the alternative choices, but even if it were, this does not invalidate what follows.)

Conclusion: the equation presented in 1. above has "delta" nonnegative, therefore Viète´s formulas may be used. (They are commonly known as "sum and product".)

3. If you realize the SUM of the roots (of the equation presented in 1. above) does not depend on the "c" coefficient (as I did), the fact that the coefficients "a" and "b" are known (1 and 3), makes it possible to guarantee that the sum of 4 and r (the second root) is -3 (= -b/a) , without finding k .

I hope all these make sense and, of course, I am absolutely comfortable to receive refutations. We are all human beings... therefore I may be wrong. I simply believe it´s not the case. (*)

BTGmoderatorDC 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

Alternate approach:

x² + 3x = 10-k
x² + 3x = constant.

Since x=4 and the correct answer choice are both roots, they must yield the same result when plugged into x² + 3x.
Plugging x=4 into x² + 3x, we get:
4² + 3*4 = 28.
If we test the answer choices, only x=-7 yields the same result when plugged into x² + 3x:
(-7)² + 3(-7) = 28.

The correct answer is A.

BTGmoderatorDC 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

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