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# IS THERE ANY OTHER WAY EXCEPT USING THE Quadratic formula

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491
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sabi92  Jul 12, 2015

#1
+27061
+10

You can factor it as:

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

so

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

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Alan  Jul 12, 2015
#1
+27061
+10

You can factor it as:

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

so

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

.

Alan  Jul 12, 2015
#2
+93691
+5

Thanks Alan,

OR

You could solve it by completing the square method

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

Melody  Jul 13, 2015