a) What are the values of #a# and #b# if #x^2+6x-3=(x+a)^2+b#? b) What is the value of #x# if #x^2+6x-3=0#?

1 Answer
Mar 4, 2017

Answer:

a)#color(white)("XX")(a,b)=(3,-12)#
b)#color(white)("XX")x=-3+-2sqrt(3)#

Explanation:

Part a
#x^2+6x-3=(x+a)^2+b#
#color(white)("XXXXXXX")=x^2+2ax+a^2+b#

#rArr#
#color(white)("XX"){:(2ax=6x,color(white)("XXX"),a^2+b=-3),(rarr a=3,,rarr 3^2+b=-3), (,,rarr 9+b=-3), (,,rarr b=-12) :}#

Part b
#x^2+6x-3=0#

There are several ways to solve this.
I will use a method called "completing the square".

First remove the constant from the variable side by adding #3# to both sides:
#color(white)("XX")x^2+6x=3#

We want to turn the left side into a squared binomial.
Remember that #(x+p)^2=x^2+2px+p^2#
So if #x^2+6x# are to be the first two terms of the expansion of a squared binomial #x^2+2px+p^2#
then #2p# must be equal to #6#
#p# must be #3#
and #p^2# must be #3^2=9#

So by adding #3^2# to both sides, we change the left side into a squared binomial
#color(white)("XX")x^2+6x+3^2=3+3^2#

Re-writing the left side as a unexpanded squared binomial
and simplifying the right side:
#color(white)("XX")=(x+3)^2=12#

Then using the square root property:
#color(white)("XX")x+3=+-sqrt(12) = +-2sqrt(3)#

and finally subtracting #3# from both sides to isolate the variable #x#
#color(white)("XX")x=-3+-2sqrt(3)#