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!!
solving for x, distribution of x...need help please
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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.
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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"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?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.
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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.
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.
the X above is divisible by the X below, so you can cancel out the two. leaving you with x^2-5x-6=0parleybaldwin wrote:"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?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.
(x-6)(x+1)=0
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x(x- {(5x+6)/x})= 0parleybaldwin wrote:"Now x *(x ^2 -5x -6)/ x = 0" - How do you get rid of the x outside the parentheses?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.
=> 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
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