1. solving quadratic equation by completing the square

HELP NEEDED

solve the following by completing the square

2x sqaured (dont know to insert power of 2) + 4x - 6

i know i have to divide it by 2 so i get

x sqaured + 2x - 3

but i dont knwo how to work it

also once i have got the answer i have to check it how do i you do this

2. Hi

$\displaystyle 2x^2 + 4x - 6 = 2(x^2 + 2x - 3) = 2(x^2 + 2x + 1 - 4) = 2 \left[(x+1)^2 - 4)\right]$

3. Originally Posted by jm16
HELP NEEDED

solve the following by completing the square

2x sqaured (dont know to insert power of 2) + 4x - 6

i know i have to divide it by 2 so i get

x sqaured + 2x - 3

but i dont knwo how to work it

also once i have got the answer i have to check it how do i you do this

I don't understand what you are asked to solve(?). I guess that you should factor the given term:

$\displaystyle 2x^2 + 4x - 6 = 2(\color{red}x^2+2x\color{black}-3)$

Use the red term to complete the square:

$\displaystyle 2(x^2+2x \bold{{\color{blue}+ 1 - 1}}-3) = 2(\underbrace{(x+1)^2-4}_{\text{difference of squares}})$

A difference of squares can be factored immediately:

$\displaystyle 2((x+1)^2-4) = 2((x+1+2)(x+1-2)) = 2(x+3)(x-1)$

4. Originally Posted by jm16
HELP NEEDED

solve the following by completing the square

2x sqaured (dont know to insert power of 2) + 4x - 6

i know i have to divide it by 2 so i get

x sqaured + 2x - 3

but i dont knwo how to work it

also once i have got the answer i have to check it how do i you do this

unless there's an "= 0" in your problem, you cannot just divide by 2 like that, youwould have to factor out a 2 in that case. the thing to note about complete squares is that the constant term is always $\displaystyle \left( \frac {\text{the coefficient of }x}2 \right)^2$. so we normally add and subtract this term (so that we effectively add 0 and hence do not change the problem) and then simplify to get the complete sqaure.

i will assume we have $\displaystyle 2x^2 + 4x - 6 = 0$

$\displaystyle \Rightarrow x^2 + 2x - 3 = 0$

$\displaystyle \Rightarrow x^2 + 2x {\color{red} + 1^2 - 1^2} - 3 = 0$

(Note that $\displaystyle x^2 + 2x + 1$ is a complete square)

$\displaystyle \Rightarrow (x + 1)^2 - 1 - 3 = 0$

$\displaystyle \Rightarrow (x + 1)^2 - 4 = 0$

can you finish?

EDIT: Pfft! too late!