GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Oct 2019, 12:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Two students were asked to solve the equation

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8017
GMAT 1: 760 Q51 V42
GPA: 3.82

### Show Tags

17 Jan 2018, 01:07
1
1
00:00

Difficulty:

45% (medium)

Question Stats:

88% (02:36) correct 13% (01:38) wrong based on 67 sessions

### HideShow timer Statistics

[GMAT math practice question]

Two students were asked to solve the equation $$x^2+px+q=0$$. Alice’s answer was $$x=1$$ and $$5$$. This answer is incorrect for the original equation, but correct for the equation $$x^2+px+r=0.$$ Bob’s answer was $$x=-2$$ and $$-4$$. This answer is incorrect for the original equation, but correct for $$x^2+sx+q=0$$. What are the solutions of the original equation?

$$A. 1, 8$$
$$B. -1,-8$$
$$C. 2,4$$
$$D. -1,-5$$
$$E. 1,-5$$

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 17 Oct 2016 Posts: 313 Location: India Concentration: Operations, Strategy GPA: 3.73 WE: Design (Real Estate) Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 01:29 1 sum of roots of x^2+px+r=0 is -p=1+5 (Alice's eqn) p=-6 product of roots of x^2+sx+q=0 is q=-2*(-4) (Bob's eqn) q=8 now the original equation is (x^2)-6x+8=0 whose roots are 2 and 4 which is option C But OA is B. Can someone explain? _________________ Help with kudos if u found the post useful. Thanks Intern Joined: 15 Jan 2018 Posts: 1 Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 11:15 I also think it is Answer C: x^2+px+r with x1=1 , x2=5 correspons to (x-1)*(x-5) = x^2-6x+5 x^2+sx+q with x1=-2 , x2=-4 correspons to (x+2)*(x+4) = x^2+6x+8 so x^2+px+q = x^2-6x+8 = 0 if you use pq-formula you will get x1=2 , x2=4 Manager Joined: 22 May 2017 Posts: 118 Re: Two students were asked to solve the equation [#permalink] ### Show Tags 17 Jan 2018, 11:25 I think there is some error in OA solved it individually as well (x-1)(x-5) = x2-6x+5 (x+2)(x+4) = x2+6x+8 therefore , x2-6x+8 correct equation roots are (x-2)(x-4) _________________ Kudos please if explanation helped ------------------------------------------------------------------------------------------------- Don't stop when you are tired , stop when you are DONE . Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Two students were asked to solve the equation [#permalink] ### Show Tags Updated on: 21 Jan 2018, 09:41 => Using the factor theorem with Alice’s solution yields $$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$ So, $$p = -6$$. Using the factor theorem with Bob’s solution yields $$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$ So, $$q = 8.$$ Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$. Therefore, the answer is C. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Originally posted by MathRevolution on 19 Jan 2018, 01:19.
Last edited by MathRevolution on 21 Jan 2018, 09:41, edited 1 time in total.
Senior Manager
Joined: 17 Oct 2016
Posts: 313
Location: India
Concentration: Operations, Strategy
GPA: 3.73
WE: Design (Real Estate)

### Show Tags

19 Jan 2018, 03:03
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
$$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$
So, $$p = -6$$.

Using the factor theorem with Bob’s solution yields
$$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$
So, $$q = 8.$$

Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$.

Please clarify what is the OA? It has to be C. in that case you have to edit the OA
_________________
Help with kudos if u found the post useful. Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 58453

### Show Tags

19 Jan 2018, 03:53
Sasindran wrote:
MathRevolution wrote:
=>

Using the factor theorem with Alice’s solution yields
$$(x-1)(x-5) = x^2 – 6x + 5 = x^2+px+r.$$
So, $$p = -6$$.

Using the factor theorem with Bob’s solution yields
$$(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.$$
So, $$q = 8.$$

Thus, the original equation is $$x^2-6x+8 = (x-2)(x-4) = 0$$. Its roots are $$2$$ and $$4$$.

Please clarify what is the OA? It has to be C. in that case you have to edit the OA

The OA is C. Edited. Thank you.
_________________
Re: Two students were asked to solve the equation   [#permalink] 19 Jan 2018, 03:53
Display posts from previous: Sort by