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6x=x^2

x=?

 Nov 17, 2014

Best Answer 

 #11
avatar+118587 
+5

Hi Sabi92, 

Welcome to the Web2 forum   

 

This is a very strange place to put your question.

Just put it on a new post next time.  :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

 

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.  

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

 

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\
=a(2b^2-c)-b(-c+2b^2)\\\\
=a\textcolor[rgb]{0,1,0}{(2b^2-c)}-b\textcolor[rgb]{0,1,0}{(2b^2-c)}\\\\
$Now you have a lots of the green stuff minus b lots of the green stuff$\\\\
$which equals (a-b) lots of the green stuff$\\\\
=(a-b)(2b^2-c)\\\\$$

 Jun 7, 2015
 #1
avatar+118587 
+5

6*x=x*x

x must be 6    

 Nov 17, 2014
 #2
avatar
0

no i have these answers x=0 and x=3 but i dont the way how to solve it

 Nov 17, 2014
 #3
avatar+26364 
+5

 i dont know how to solve can someone help me

6x=x^2

x=?

$$\small{\text{
$
\begin{array}{rcl}
6x & = & x^2 \\
x^2 - 6x &=& 0 \\
\underbrace{x}_{=0}(\underbrace{x-6}_{=0}) &=& 0 \\
\end{array}
$
}}$$

1. x = 0

2. x-6 = 0 | +6

    x    = 6

 Nov 17, 2014
 #4
avatar
0

thank you very much:)!

 Nov 17, 2014
 #5
avatar+118587 
0

Very nice Heureka :)

 Nov 17, 2014
 #6
avatar
0

oops im sorry i wrote a wrong question the real question is 

6x=2x^2

x=?

 

so how can i solve it? 

 Nov 17, 2014
 #7
avatar+26364 
+5

 i dont know how to solve can someone help me

6x=2x^2

x=?

$$\small{\text{
$
\begin{array}{rcl}
6x & = & 2x^2 \\
2x^2 - 6x &=& 0 \\
\underbrace{x}_{=0}(\underbrace{2x-6}_{=0}) &=& 0 \\
\end{array}
$
}}$$

1. x = 0

2. 2x-6 = 0 | +6

    2x    = 6 | : 2

     x    = 6 /2

     x    = 3

 Nov 17, 2014
 #8
avatar+118587 
+5

It is similar to what Heureka did.

 

$$\\6x=2x^2\\
3x^2-6x=0\\
3x(x-2)=0$$

you should be able to finish it :)

 Nov 17, 2014
 #9
avatar
0

ok thnk you but its 2x(x-3)=0

 Nov 17, 2014
 #10
avatar+262 
0

i need to do a factorization but i dont get the right answer.help please.

 

a(2b^2-c)+b(c-2b^2)

that's what i did:

2ab^2-ac+bc-2b^3=a(2b^2-c)+b(c-2b^2)

now what?

 Jun 7, 2015
 #11
avatar+118587 
+5
Best Answer

Hi Sabi92, 

Welcome to the Web2 forum   

 

This is a very strange place to put your question.

Just put it on a new post next time.  :)

I really like that fact that you have had a go at it yourself and that you have shown us your work :)

 

$$\\a(2b^2-c)+b(c-2b^2)$$

what you need to notice here is that the brackets contain the same things except the signs are different.  

I am going to look at the second bracket. I could just as easily chosen the first bracket :)

 

$$c-2b^2 = -1(-c+2b^2)=-1(2b^2-c)$$

so

$$+b(c-2b^2)=+b*-1(2b^2-c) = -b(2b^2-c)$$

so

$$\\a(2b^2-c)+b(c-2b^2)\\\\
=a(2b^2-c)-b(-c+2b^2)\\\\
=a\textcolor[rgb]{0,1,0}{(2b^2-c)}-b\textcolor[rgb]{0,1,0}{(2b^2-c)}\\\\
$Now you have a lots of the green stuff minus b lots of the green stuff$\\\\
$which equals (a-b) lots of the green stuff$\\\\
=(a-b)(2b^2-c)\\\\$$

Melody Jun 7, 2015

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