Completing the Square - Math Help Needed!

[psycoswimmer]

So, I've been trying to figure out this problem for a while and I can't figure out how to do it. The question states: "Solve 2x^2 - 12x + 4 = 0 by completing the square, expressing the result in simplest radical form." I'm not sure how to continue with completing the square with a leading coefficient... I tried pulling out a 2 but then got confused up to the point were I had (2)(x+3)^2=7.

Any help is greatly appreciated!

 

[AngryApple]

I don't know if I did it right:

x = 3 + root 7

 

[fireshot91]

My first bet would be to factor it, making it 2(x^2-6x+2)=0, but that leads you nowhere, since you can't factor that anymore.

 

[sammich]

Your first step is to get the coeff of x^2 to be equal to 1

Stepwise:

.x² - 6x + 2 = 0............after dividing by 2
(x² - 6x + 9) + 2 . 9 = 0...complete the square: add (b/2)² = 9
(x - 3)² = 7................solving for x
.x - 3...= √7
.x.......= -3 - √7

 

[imaketouchtheme]

Ok, when there's a coefficient in front of the x^2, you have to first divide the entire equation by the coefficient, which will look like this:

x^2-6x+2=0

All I did was divide everything by 2.

After that, move your "c" term over (since the equation is in the form of ax^2-bx+c=0)

Now you have x^2-6x=-2

If you need anymore help from there, let me know.

 

[Tomorrow]

Your first step is to get the coeff of x^2 to be equal to 1

Stepwise:

.x² - 6x + 2 = 0............after dividing by 2
(x² - 6x + 9) - 7 = 0.......complete the square: add (b/2)² = 9
(x - 3)² = 7................solving for x
.x - 3...= ±√7
.x.......= 3 ± √7

Fixed. You were close. Remember, a quadratic equation can have two roots, often related to the positive and negative roots of (b^2-4ac). It's a good idea to check your work using the quadratic formula.

 

[sammich]

Fixed. You were close. Remember, a quadratic equation can have two roots, often related to the positive and negative roots of (b^2-4ac). It's a good idea to check your work using the quadratic formula.

Oh man, bad memories of little mistakes in maths exams.

Thanks for fixing those up for me. And to think, I'm studying engineering

 

[Tomorrow]

Oh man, bad memories of little mistakes in maths exams.

Thanks for fixing those up for me. And to think, I'm studying engineering

You poor soul. (spoken by a 40-year-old engineer)

 

[psycoswimmer]

Thanks everyone, I finally got it. And yes, the answer is 3 ± √7.