Linear equation

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Mon Nov 17, 2014 4:45 am

by [email protected] » Sat May 09, 2015 7:15 am

Hello,

I am blanking out on this question. I got it before, but now I can;t figure out how I went through the steps to cancel the x's.

x(x-5x+6/x) =0

Looking for step by step approach.

Thanks!

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Source: Beat The GMAT — Problem Solving | Sat May 09, 2015 7:15 am

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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 » Sat May 09, 2015 7:49 am

[email protected] wrote:Hello,

I am blanking out on this question. I got it before, but now I can;t figure out how I went through the steps to cancel the x's.

x(x-5x+6/x) =0

Looking for step by step approach.

Thanks!

Are you sure the equation is x(x- 5x + 6/x)?
I have a feeling it's supposed to be x(x - 5 + 6/x)
This is the equation I'll solve.

General idea: If (something)(other thing) = 0, then something = 0 or other thing = 0

So, if x(x - 5 + 6/x) = 0, then x = 0 or x - 5 + 6/x = 0
So, IT SEEMS that we now have at least one solution: x = 0
HOWEVER, in the second part of the equation, we have 6/x. Here, if x = 0, then 6/x is undefined.
So, x cannot equal zero.

What about the equation x - 5 + 6/x = 0?
Let's first eliminate the fraction by multiplying both sides by x to get: xÂ² - 5x + 6 = 0
Factor to get: (x - 2)(x - 3) = 0
This means that x - 2 = 0 or x - 3 = 3
If x - 2 = 0, then x = 2
If x - 3 = 0, then x = 3

So, there are 2 possible values of x that satisfy the equation: x = 2 and x = 3

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Mon May 11, 2015 6:17 am, edited 1 time in total.

Brent Hanneson - Creator of GMATPrepNow.com

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Sat May 09, 2015 7:52 am

[email protected] wrote:Hello,

I am blanking out on this question. I got it before, but now I can;t figure out how I went through the steps to cancel the x's.

x(x-5x+6/x) =0

Looking for step by step approach.

Thanks!

Hello [email protected],

x=0 will not be a solution to this quadratic,since, division by zero is not admissible.Now, simplify as follows:-
x(-4x^2 + 6)/x=0
The x in the numerator and denominator gets cancelled, so that we are left with:-4x^2+ 6=0
or, 4x^2=6
or, x^2=3/2
or, x= +/-(3/2)^1/2
Check, both the roots obtained, by plugging in the original equation. Both roots are valid!

800. Arjun's-Bird-Eye

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[email protected]: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Mon Nov 17, 2014 4:45 am

by [email protected] » Sun May 10, 2015 7:19 am

You are correct, Brent! Sorry - I added an extra x...thanks for correcting and answering the question. I'm clear now:)

Thanks again,
B

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Aman verma: Master | Next Rank: 500 Posts; Posts: 304; Joined: Wed Jan 27, 2010 8:35 amFollowed by: 3 members; Location: International Space Station; Thanked: 11 times

by Aman verma » Mon May 11, 2015 1:27 am

[email protected] wrote:You are correct, Brent! Sorry - I added an extra x...thanks for correcting and answering the question. I'm clear now:)

Thanks again,
B

Even if we remove the extra x, x=0 cannot be a valid solution for the quadratic,since division by zero is undefined. After removing the extra x, we have:-
x(x-5+6/x)=0
or, (x-2)(x-3)=0
or, x=3 or x=2
So, there are 2 possible values of x that satisfy the equation: x=2 and x=3.
Zero is not a valid solution in this equation!

800. Arjun's-Bird-Eye

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GMAT/MBA Expert

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 » Mon May 11, 2015 6:17 am

Aman verma wrote:
[email protected] wrote:You are correct, Brent! Sorry - I added an extra x...thanks for correcting and answering the question. I'm clear now:)

Thanks again,
B
Even if we remove the extra x, x=0 cannot be a valid solution for the quadratic,since division by zero is undefined. After removing the extra x, we have:-
x(x-5+6/x)=0
or, (x-2)(x-3)=0
or, x=3 or x=2
So, there are 2 possible values of x that satisfy the equation: x=2 and x=3.
Zero is not a valid solution in this equation!

Good catch, Aman.
I edited my response accordingly.

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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