Guide 1 p. 91 #9 - keep getting same (wrong) answer

[official_jj]

Okay the question is:

Solve for x:
x (x - ((5x+6)/x)) = 0

I am okay up to:
x^2 - 5x - 6 = 0

However, this is what I'm doing next:

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

x = 6, x = 11

At the same step, the book does this:

(x-6)(x+1) = 0

x = 6, x = -1

I understand that they "reverse foiled" to get (x-6)(x+1), but I am confused why solving it the way I did is giving me a different answer. Perhaps I'm just forgetting some basic algebra rule, but I thought that if a variable is a common factor, it can be removed from the equation and then you solve it by making what's outside the parenthesis 0 [then solving for x] and then what's inside the parenthesis 0 [then solving for x].

[akhp77]

You are trying to cheat algebraic equation

The one, which is given in your book, is correct.
Now correct your method

x (x - 5) = 6 = 6*1 = -1* -6
x = 6, -1

[tim]

It looks as though you may have misapplied a rule. When you got to

x(x-5)=6

you probably said too yourself that one of them had to equal 6, so you got x=6 or x-5=6. This only works when the right side is 0. If two things multiply to be 0, one of them HAS to be 0. Not true for any other number. That is why it is absolutely essential in problems of this type to get one side to equal 0 and factor the other side.

[pmclose]

I'm having trouble completing the quadratic x^2(-5x-6)=0. Can someone please explain how you get the x^2 portion of the quadratic? In other words, I'm not sure how to cancel out or isolate the bottom part of the fraction (x-(5x+6/x))=0 which leaves x^2(-5x-6)=0 . After that, I'm able to complete the problem correctly. Thanks in advance.

[Ben Ku]

I'm having trouble completing the quadratic x^2(-5x-6)=0. Can someone please explain how you get the x^2 portion of the quadratic? In other words, I'm not sure how to cancel out or isolate the bottom part of the fraction (x-(5x+6/x))=0 which leaves x^2(-5x-6)=0 . After that, I'm able to complete the problem correctly. Thanks in advance.

The question in the book is:
x[x - ((5x+6)/x))] = 0

We can distribute the outer "x" to each term inside using the distributive property:
a(b - c) = ab - ac
x^2 - x((5x+6)/x)) = 0

Note that in the second term, there is an "x" in the numerator and denominator, so they can cancel out. Now you get
x^2 - (5x + 6) = 0
x^2 - 5x - 6 = 0

Hope that helps.