SOLUTION: (x-6x sq root x)=0 Solve for x
In response to an inquiry -I was told to "use the conjugate". After yet another inquiry, I was given:
(x-6x sq root x)(x+6x sq root x)=0(x+6x sq
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-> SOLUTION: (x-6x sq root x)=0 Solve for x
In response to an inquiry -I was told to "use the conjugate". After yet another inquiry, I was given:
(x-6x sq root x)(x+6x sq root x)=0(x+6x sq
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Question 146097: (x-6x sq root x)=0 Solve for x
In response to an inquiry -I was told to "use the conjugate". After yet another inquiry, I was given:
(x-6x sq root x)(x+6x sq root x)=0(x+6x sq root x)
= x squared -36xsquaredx and told to solve by factoring it out. I understand the conjugate part (kind of) but I've never seen -36xsquaredx! How do I factor this out?
By the way, I know the answer is {0,1/36} and I still can't figure it out.
This is an online course. I have searched the book and cannot find a sample of this problem anywhere. Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! x-6x*sqrt(x)=0
-6x*sqrt(x)=-x
6x*sqrt(x)=x
36x^2*x=x^2 square each side
36x^3=x^2
36x^3-x^2=0
x^2(36x-1)=0
x=0
36x=1
x=1/36
.
You also can do it the way suggested:
(x-6x*sqrt(x))*(x+6x*sqrt(x))=(x+6x*sqrt(x)*0 Multiply each side by x+6x*sqrt(x)
x^2-36x^3=0 This is the same as (x-y)(x+y)=x^2-y^2. Let y=6x*sqrt(x)
36x^3-x^2=0 Same as line 6 above.
.
Ed
