How do you solve #x^2 + 3x +6 =0# by completing the square?
2 Answers
# x = ± sqrt3 - 3 #
Explanation:
By adding 9 to both sides of the equation to obtain:
#( x^2 + 3x + 9 )+ 6 = 9# now
#(x + 3 )^ 2 = 9-6 = 3#
# ( x + 3 )^2 color(black)( " is a perfect square ") # Taking the 'square root' of both sides :
# sqrt((x+ 3 )^2 )= sqrt3# hence x + 3 = ±
#sqrt3 # so x = ±
#sqrt3 - 3 #
there are two
Explanation:
Completing the square method:
Do this only when the numerical coefficient of
Start with the numerical coefficient of
Divide this number by 2 then square the result. That is
Add
the first three terms now become one group which is a PST-Perfect Square Trinomial
Finally, transpose the
take note:
therefore
there are two