• Oct 14th 2009, 05:25 AM
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?
• Oct 14th 2009, 05:33 AM
craig
Quote:

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 $\displaystyle x^2 + 2bx +c = 0$, 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
• Oct 14th 2009, 05:34 AM
masters
Quote:

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.

$\displaystyle x^2+6x=-1$

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?