How do you solve #x(x+6)=-2# using the quadratic formula?

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Answer 1 · 2017-10-29T11:58:42

# x= -0.354 #
OR
# x = -5.645#

#x(x+6)=-2#

Writing it in standard form:

#x^2 +6x+2=0#

Compare with equation #ax^2 +bx+c=0#

#a = 1#
#b=6#
#c=2#

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

#x=-0.354248688~~-0.354#---truncated and approximate value.

Or

#x =-5.645751311~~-5.645#---truncated and approximate value.

#therefore x= -0.354 #
OR
# x = -5.645#

Answer 2 · 2017-10-29T12:05:32

#x = -3 + root2 7 or approx -0.3542... #
#x = -3 - root2 7 or approx -5.6457... #

Expand and rearnage;
#x^2 + 6x = -2 #
#x^2 + 6x + 2 = 0#

Then apply to the quadratic formula;
# a =1, b = 6, c=2#

#x = (-6+root2 ( 6^2 - ( 4*1*2) ))/2#
or #x = (-6-root2 ( 6^2 - ( 4*1*2) ))/2#

To give
#x = (-6+root2 ( 28 ))/2#
or #x = (-6-root2 ( 28 ))/2#

Hence
#x = (-6+2root2 ( 7 ))/2#
or #x = (-6-2root2 ( 7 ))/2#

Finally giving;
#x = -3 + root2 7 or approx -0.3542... #
#x = -3 - root2 7 or approx -5.6457... #