How do you solve the quadratic #3+5/(2x)=1/x^2# using any method?

1 Answer
Sep 21, 2016

Answer:

#x =0.295 " or "x=-1.129#

Explanation:

If you have an equation which has fractions, you can get rid of them by multiplying each term by the LCM of the denominators.

In the equation #" "3+5/(2x)=1/x^2" the LCD is " color(blue)(2x^2)#

#color(white)(x)color(blue)(2x^2)xx3+color(blue)(cancel(2x^2)^x)xx5/(cancel(2x))=color(blue)(2cancelx^2)xx1/cancelx^2#

#color(white)(xxxxxxxxxxx)6x^2+5x=2" "larr# no fractions!

#6x^2 +5x -2 = 0" "rarr# the options are: #"find factors"#
#color(white)(xxxxxxxxxxxxxxxxx.xxxxxxxxx)"complete the square"#
#color(white)(xxxxxxxxxxxxxxxxx.xxxxxxxxx)"quadratic formula"#

This quadratic expression does not factor, use the formula.

#x = (-b +-sqrt(b^2 -4ac))/(2a)#

#x = (-5 +-sqrt(5^2 -4(6)(-2)))/(2(6))#

#x = (-5 +-sqrt(25 +48))/(12)#

#x = (-5 +sqrt(73))/(12)" or "x = (-5 +sqrt(73))/(12)#

#x =0.295 " or "x=-1.129#