Question

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`?

Answer 1

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)`