How do you solve by using the quadratic formula: #x^2 + 7x = -3#?

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Answer 1 · 2016-05-09T19:22:19

#x=-7/2+-sqrt(37)/2#

First add #3# to both sides to get:

#x^2+7x+3 = 0#

This is in the form #ax^2+bx+c = 0# with #a=1#, #b=7# and #c=3#.

The roots are given by the quadratic formula:

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

#=(-7+-sqrt(7^2-(4*1*3)))/(2*1)#

#=(-7+-sqrt(49-12))/2#

#=(-7+-sqrt(37))/2#

#=-7/2+-sqrt(37)/2#

Note that since #37# is prime, the square root does not simplify further.

#sqrt(37)# does have a simple continued fraction expansion which you can truncate to give rational approximations:

#sqrt(37) = [6;bar(12)] = 6+1/(12+1/(12+1/(12+1/(12+1/(12+...)))))#