# 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?

Mar 4, 2017

a)$\textcolor{w h i t e}{\text{XX}} \left(a , b\right) = \left(3 , - 12\right)$
b)$\textcolor{w h i t e}{\text{XX}} x = - 3 \pm 2 \sqrt{3}$

#### Explanation:

Part a
${x}^{2} + 6 x - 3 = {\left(x + a\right)}^{2} + b$
$\textcolor{w h i t e}{\text{XXXXXXX}} = {x}^{2} + 2 a x + {a}^{2} + b$

$\Rightarrow$
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
${x}^{2} + 6 x - 3 = 0$

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:
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 6 x = 3$

We want to turn the left side into a squared binomial.
Remember that ${\left(x + p\right)}^{2} = {x}^{2} + 2 p x + {p}^{2}$
So if ${x}^{2} + 6 x$ are to be the first two terms of the expansion of a squared binomial ${x}^{2} + 2 p x + {p}^{2}$
then $2 p$ 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
$\textcolor{w h i t e}{\text{XX}} {x}^{2} + 6 x + {3}^{2} = 3 + {3}^{2}$

Re-writing the left side as a unexpanded squared binomial
and simplifying the right side:
$\textcolor{w h i t e}{\text{XX}} = {\left(x + 3\right)}^{2} = 12$

Then using the square root property:
$\textcolor{w h i t e}{\text{XX}} x + 3 = \pm \sqrt{12} = \pm 2 \sqrt{3}$

and finally subtracting $3$ from both sides to isolate the variable $x$
$\textcolor{w h i t e}{\text{XX}} x = - 3 \pm 2 \sqrt{3}$