a) What are the values of aa and bb if x^2+6x-3=(x+a)^2+bx2+6x3=(x+a)2+b? b) What is the value of xx if x^2+6x-3=0x2+6x3=0?

1 Answer
Mar 4, 2017

a)color(white)("XX")(a,b)=(3,-12)XX(a,b)=(3,12)
b)color(white)("XX")x=-3+-2sqrt(3)XXx=3±23

Explanation:

Part a
x^2+6x-3=(x+a)^2+bx2+6x3=(x+a)2+b
color(white)("XXXXXXX")=x^2+2ax+a^2+bXXXXXXX=x2+2ax+a2+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)