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common root

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Source: — Problem Solving |

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by Brent@GMATPrepNow » Tue Aug 07, 2012 4:33 pm
grandh01 wrote:Which of the following equations has
a root in common with x^2 - 6x + 5 = 0?
(A) x^2 + 1 = 0
(B) x^2 - x - 2 =0
(C) x^2 - 10x - 5 =0
(D) 2x^2 - 2 =0
(E) x^2 - 2x - 3 =0
Step 1: Solve the given equation: x^2 - 6x + 5 = 0
This is a quadratic set equal to zero, so let's factor to get:
(x-1)(x-5)=0
So, we have two solutions (roots): x=1 or x=5

Step 2: Solve the other 5 equations to see which one has a root (solution) of x=1 or x=5

IMPORTANT: It appears that the only way to answer this question is to keep checking every single answer choice until we find that one that has a solution of either x=1 or x=5. Given this, where do you think the test-maker would hide the correct answer? In these situations, I always start at E and work my way up. Is the answer to these questions always E (or perhaps D)? No, but it's more likely that the correct answer is near the bottom.

Okay, E: x^2 - 2x - 3 =0
Factor to get: (x-3)(x+1)=0
So, x=3 or x=-1
No shared solutions (roots) so keep moving.

D: 2x^2 - 2 =0
Factor: 2(x^2 - 1) = 0
Keep factoring: 2(x+1)(x-1)=0
So, x=1 or x=-1

We have a common solution, so the correct answer must be D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by adthedaddy » Thu Aug 09, 2012 8:19 am
To add to the following,

The roots for given equation are x=1, x=5

For the options, the roots are as follows;
A: x^2=-1
B: x=2, x=-1
C: Complex roots
D: x=1
E: x=-1, x=3

Thus, ans is D
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by gary391 » Thu Jul 07, 2016 7:05 pm
For a quadratic equation: ax2 + bx + C = 0. A real number x will be called a solution or root if it satisfies the equation. Since for the given problem, we can calculate the roots as x = 1 and x = 5. Now, we are asked to identify which of the options will have the common root. Therefore, by just plugging in x = 1 and X = 5 in each of the options to identify the satisfied equation. Answer is D
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by OptimusPrep » Fri Jul 08, 2016 12:57 am
grandh01 wrote:Which of the following equations has
a root in common with x2 - 6x + 5 = 0?
(A) x2 + 1 = 0
(B) x2 - x - 2 =0
(C) x2 - 10x - 5 =0
(D) 2x2 - 2 =0
(E) x2 - 2x - 3 =0
x^2 - 6x + 5 = 0
(x-5)(x-1) = 0
The roots are 1 and 5

We need to check for each option to calculate the roots. This can be done by plugging in the values in the equations

(A) x2 + 1 = 0. None of 1 or 5 satisfies this equation
(B) x2 - x - 2 =0. None of 1 or 5 satisfies this equation
(C) x2 - 10x - 5 =0. None of 1 or 5 satisfies this equation
(D) 2x2 - 2 =0. x = 1 satisfies this equation. Hence this has a common root
(E) x2 - 2x - 3 =0. None of 1 or 5 satisfies this equation

Correct Option: D
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by Matt@VeritasPrep » Fri Jul 08, 2016 1:54 pm
Just out of curiosity, are people the ones resurrecting these threads, or is somebody testing out a script/bot that's written to answer forum questions?
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