+0  
 
0
546
2
avatar+262 

 Jul 12, 2015

Best Answer 

 #1
avatar+27396 
+10

You can factor it as:

 

$$(x-\frac{1}{2})(x-4m+\frac{1}{2})=0$$

 

so  

$$\\ x=\frac{1}{2}\\ or\\x=4m-\frac{1}{2}$$

.

 Jul 12, 2015
 #1
avatar+27396 
+10
Best Answer

You can factor it as:

 

$$(x-\frac{1}{2})(x-4m+\frac{1}{2})=0$$

 

so  

$$\\ x=\frac{1}{2}\\ or\\x=4m-\frac{1}{2}$$

.

Alan Jul 12, 2015
 #2
avatar+95369 
+5

Thanks Alan,

OR

You could solve it by completing the square method

 

$$\\x^2-4xm+(2m-\frac{1}{4})=0\\\\
x^2-4xm\;=\;-(2m-\frac{1}{4})\\\\
x^2-4xm+(2m)^2\;=\;-2m+\frac{1}{4}+(2m)^2\\\\
(x-2m)^2\;=\;-2m+\frac{1}{4}+4m^2\\\\
(x-2m)^2\;=\;4m^2-2m+\frac{1}{4}\\\\
(x-2m)^2\;=\;4(m^2-\frac{m}{2}+\frac{1}{16})\\\\
(x-2m)^2\;=\;4(m-\frac{1}{4})^2\\\\
x-2m\;=\;\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m\pm 2(m-\frac{1}{4})\\\\
x\;=\;2m+ 2m-\frac{1}{2}\qquad or \qquad x\;=\;2m- 2m+\frac{1}{2}\\\\
x\;=\;4m-\frac{1}{2}\qquad or \qquad x\;=\;\frac{1}{2}\\\\$$

.
 Jul 13, 2015

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