How do you find the roots of #.0625x^2=.125x+15# by factoring?

1 Answer
Feb 24, 2017

Answer:

#x=1+-sqrt241#

Explanation:

#0.0625x^2=0.125x+15#

A good idea would be to remove the decimal coefficients. The smallest multiplicand which gives whole number coefficients is #16#, so we'll multiply the equation by #16#.

#x^2=2x+240#

#x^2-2x-240=0#

we can't factorise this equation so we're going to have to use the quadratic formula:

#x=(2+-sqrt(2^2-4(-240)(1)))/2=1+-sqrt(241)#