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solving for x, distribution of x...need help please

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by zander21 » Tue Oct 25, 2011 12:07 pm
solve for x:

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

the solution says to distribute the multiplication of x. note, when you cancel the x in the denominator, the quantity 5x+6 is implicitly enclosed in parentheses.

x^2 - (5x+6) = 0
x^2 - 5x - 6 = 0 and then of course this is easy it's 6 or -1

BUT I can't figure out how to get rid of the two x's in the original parentheses...how do I distribute the multiplication of x? The source is page 91 (solution on 94) of MGMAT strategy guide Number Properties.

Any help would be appreciated thanks!!
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by samyukta » Tue Oct 25, 2011 12:24 pm
x(x- (5x+6/x) ) = 0.




Should have been written as : x(x- {(5x+6)/x)} ) = 0.

Now x *(x ^2 -5x -6)/ x = 0

we can simplify it to (x ^2 -5x -6) = 0 which gives soln as x = 6 or -1 basic quadriatic eqn.
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by zander21 » Tue Oct 25, 2011 12:34 pm
samyukta wrote:x(x- (5x+6/x) ) = 0.




Should have been written as : x(x- {(5x+6)/x)} ) = 0.

Now x *(x ^2 -5x -6)/ x = 0

we can simplify it to (x ^2 -5x -6) = 0 which gives soln as x = 6 or -1 basic quadriatic eqn.
"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?
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by suyashgupta0562 » Tue Oct 25, 2011 5:06 pm
Hey,

If we solve the expression in the inner parentheses (5x+6/x) first then we get (5x^2+6)/x taking x as the LCM

Then we subtract the above expression from x.

x-(5x^2+6)/x = (x^2-5x^2-6)/x

Multiplying this with x now, the x in the denominator cancels out giving us (x^2-5x^2-6) =0

Which simplifies to -4x^2 - 6 = 0

x^2 = -6/4 which would make x imaginary.
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by rooster » Tue Oct 25, 2011 9:29 pm
parleybaldwin wrote:
samyukta wrote:x(x- (5x+6/x) ) = 0.




Should have been written as : x(x- {(5x+6)/x)} ) = 0.

Now x *(x ^2 -5x -6)/ x = 0

we can simplify it to (x ^2 -5x -6) = 0 which gives soln as x = 6 or -1 basic quadriatic eqn.
"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?
the X above is divisible by the X below, so you can cancel out the two. leaving you with x^2-5x-6=0

(x-6)(x+1)=0
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by Amiable Scholar » Fri Oct 28, 2011 8:58 am
parleybaldwin wrote:
samyukta wrote:x(x- (5x+6/x) ) = 0.




Should have been written as : x(x- {(5x+6)/x)} ) = 0.

Now x *(x ^2 -5x -6)/ x = 0

we can simplify it to (x ^2 -5x -6) = 0 which gives soln as x = 6 or -1 basic quadriatic eqn.
"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?
x(x- {(5x+6)/x})= 0
=> either x = 0 or x- {(5x+6)/x} =0
x= 0 not possible because x is in denominator in second expression
so
x- {(5x+6)/x} = 0 only ...
if that solves your query of removing external x :)
Amiable Scholar

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