Linear equation

[[email protected]]

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!

[Brent@GMATPrepNow]

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: x² - 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

[Aman verma]

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!

[[email protected]]

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

Thanks again,
B

[Aman verma]

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!

[Brent@GMATPrepNow]

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