"If 4 Is One of the Solutions to x² + 3x + k = 10..." — GMAT® Worked Solution

Source: Official Guide for GMAT® Review, 11th Edition

If 4 is one of the solutions to 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

Try it before reading on.

Route 1: A Given Solution Is a Given Factor

The usual instinct with a quadratic is to set it equal to zero and factor. This question lets you start halfway there. If $x = 4$ is a solution, then $x - 4$ is one of the two factors, because a solution is exactly the value that makes its factor equal zero.

So the factored form has one binomial filled in already:

$(x - 4)(x + \underline{\quad})$

The unfilled binomial comes from the coefficient of $x$. When the two binomials multiply out, the two constants, here $-4$ and whatever fills the blank, have to sum to the coefficient on the $3x$ term. That holds no matter what $k$ is, which is why $k$ can sit this out entirely: the value of the constant never touches the $x$ coefficient.

What number adds to $-4$ to make 3?

$-4 + 7 = 3$

So the full factored form is $(x - 4)(x + 7) = 0$, and the other solution is what makes the second factor zero.

The answer is (A).

Route 2: Solve for k, Then Factor

If the reasoning above felt too abstract, the concrete route is plug and chug, and it works because the problem hands you a value that makes the equation true. Substitute $x = 4$:

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

$16 + 12 + k = 10$

$28 + k = 10$

$k = -18$

Now the equation is fully known: $x^2 + 3x - 18 = 10$. Move the 10 to the left side:

$x^2 + 3x - 28 = 0$

Factor it the standard way, with two numbers that multiply to $-28$ and sum to $3$. Searching the factor pairs of 28, the candidates are 4 and 7, and the signs that produce a sum of 3 are $-4$ and $+7$:

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

The solutions are 4 and $-7$, and since 4 is already accounted for:

The answer is (A).

Choosing Between the Routes

Route 2 is concrete and hard to get lost in, and it's the better recommendation for anyone still building comfort with factoring. Route 1 gets to the same answer in a fraction of the steps, and fewer steps means fewer chances to slip, which matters on a section with no partial credit.

There's no universal answer to which one to use, and there doesn't need to be. A question this short is a good chance to run both during review and note which one you'd trust under time pressure, then let your own results set the default. Both routes stay worth knowing, because the fallback is only a fallback if it was learned.

Why This Problem Matters

The design of this question rewards reading what's given before computing. The given root is a finished factor, and the given coefficient finishes the other one; the constant $k$ is scenery. Test takers who reach for the full algebra first still get there, but they spend steps on information the question never needed, and each step is a chance for a slip.

That reading habit, asking what a given fact already buys before doing work to re-derive it, shows up across the quant section. Variables in answer choices make plugging in available. A stated rate makes a chart row solvable. A stated root makes a factor solvable. The given is usually an offer, and it costs nothing to check what it covers before declining it.

Want the full strategy? Read: GMAT® Percents, Divisibility, and Quadratics: Translate First, Solve Second

From Episode 11 of Real GMAT® Problems (The GMAT® Strategy Podcast).