If 4 is one solution of the equation x^2 + 3x + k = 10,

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BTGmoderatorDC: Moderator; Posts: 7187; Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members

by BTGmoderatorDC » Wed Aug 29, 2018 7:07 pm

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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

OA A

Source: Official Guide

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Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 amFollowed by: 34 members; Location: Grand Central / New York; Thanked: 470 times

by Jay@ManhattanReview » Wed Aug 29, 2018 9:52 pm

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

OA A

Source: Official Guide

Plugging-in the value of x = 4 in the equation x^2 + 3x + k = 10, we get

4^2 + 3*4 + k = 10
k = -18

Thus, the equation is x^2 + 3x - 18 = 10 => x^2 + 3x - 28 = 0

Factoring x^2 + 3x - 28 = 0, we get

x^2 + 7x - 4x - 28 = 0

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

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

x = 4 or -7.

The other solution is -7.

The correct answer: A

Hope this helps!

-Jay
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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

If 4 is one solution of the equation x^2 + 3x + k

by fskilnik@GMATH » Thu Aug 30, 2018 5:37 am

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 ViÃ¨teÂ´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 itÂ´s not the case. (*)

Last edited by fskilnik@GMATH on Thu Aug 30, 2018 7:20 am, edited 2 times in total.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Thu Aug 30, 2018 5:43 am

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

OA A

Source: Official Guide

Let's first determine the value of k.

Since x = 4 is a solution to the equation xÂ² + 3x + k = 10, we know that x = 4 SATISFIES the equation.
That is: 4Â² + 3(4) + k = 10
Evaluate to get: 16 + 12 + k = 10
Solve for k to get: k = -18

So, the ORIGINAL equation is xÂ² + 3x + (-18) = 10
This is the same as: xÂ² + 3x - 18 = 10
We now need to solve this equation.

First, set it equal to zero: xÂ² + 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

Brent Hanneson - Creator of GMATPrepNow.com

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Aug 30, 2018 6:31 am

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:

xÂ² + 3x = 10-k
xÂ² + 3x = constant.

Since x=4 and the correct answer choice are both roots, they must yield the same result when plugged into xÂ² + 3x.
Plugging x=4 into xÂ² + 3x, we get:
4Â² + 3*4 = 28.
If we test the answer choices, only x=-7 yields the same result when plugged into xÂ² + 3x:
(-7)Â² + 3(-7) = 28.

The correct answer is A.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

by Jeff@TargetTestPrep » Tue Sep 04, 2018 3:52 am

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

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