I'm finding it difficult to solve this question
It's solving quadratic eqaution by completing the square
x^2+6x+1=0
Any help?
Here's how I'd proceed.
When solving the square of a quadratic in the form $\displaystyle x^2 + 2bx +c = 0$, first you put into brackets, $\displaystyle (x+b)^2 + c = 0$.
So in the case of your question you get $\displaystyle (x+3)^2 +1 = 0$. However, if you multiply out $\displaystyle (x+3)^2$,you find that you get an extra 9 in your equation, to compensate for this you also place a -9 in your equation, giving you $\displaystyle (x+3)^2 - 9 +1 = 0$.
A little bit of rearranging should give you $\displaystyle (x+3)^2 = 8$, which in turn leads to $\displaystyle x+3 = \pm2\sqrt{2}$.
I presume you can continue from here to find the two solutions.
Craig
Hi Mist,
Step 1: Transpose the constant to the right hand side.
$\displaystyle x^2+6x=-1$
Step 2: Take half the coefficent of the linear term (x), square it, and add the result to both sides of the equation.
$\displaystyle x^2+6x+9=-1+9$
Step 3: You have just made the left side a perfect square trinomial.
$\displaystyle (x+3)^2=8$
Step 4: Take the square root of both sides.
$\displaystyle x+3=\pm \sqrt{8}$
I'll bet you can finish up, can't you?