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?

2. Originally Posted by Mist
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 $x^2 + 2bx +c = 0$, first you put into brackets, $(x+b)^2 + c = 0$.

So in the case of your question you get $(x+3)^2 +1 = 0$. However, if you multiply out $(x+3)^2$,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 $(x+3)^2 - 9 +1 = 0$.

A little bit of rearranging should give you $(x+3)^2 = 8$, which in turn leads to $x+3 = \pm2\sqrt{2}$.

I presume you can continue from here to find the two solutions.

Craig

3. Originally Posted by Mist
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?
Hi Mist,

Step 1: Transpose the constant to the right hand side.

$x^2+6x=-1$

Step 2: Take half the coefficent of the linear term (x), square it, and add the result to both sides of the equation.

$x^2+6x+9=-1+9$

Step 3: You have just made the left side a perfect square trinomial.

$(x+3)^2=8$

Step 4: Take the square root of both sides.

$x+3=\pm \sqrt{8}$

I'll bet you can finish up, can't you?