What steps do I take to transform x^2 + 4x + 3 = 0 into the form (x - p)^2 = q?

Question

mathslover · Answer

Use the method of Square Completion here.

johnweldon1993 · Answer

Complete the square

mathslover · Answer

$x^2 + 2(2)x + 4-4 + 3 =0 $
$(x^2 + 2(2)x + 4 ) + 3 - 4= 0$ 

What I did is that, I assumed 4x as the term equivalent to : 2xp and so I made it like : 2(2)(x) 

So, we have , x =x and p = 2 

We know the formula : $(x-p)^2 = x^2 - 2xp + p^2 $ 

We have, x^2 and 2xp but we require p^2

So, we add and subtract 2^2 in the LHS

That is : (the first step in the first Equation)

mathslover · Answer

Can you do it from here?

johnweldon1993 · Answer

Step 1) Subtract 3 from both sides.
Step 2) Take the coefficient of the 'x' ..divide it by 2....square the result...and add to both sides of the equation

$$\large x^2+4x=−3$$


4/2 = 2 ...2^2 = 4...so we have

$$\large x^2+4x+4=−3+4$$

Factor the left hand side and simplify the right hand side ...what do you get?

kewlgeek555 · Answer

I am a bit confused @mathslover  . (>.<)

Wasn't the Square Completion method where first you move the loose number to the other side of the equation, take half of the x-term and square it and then add the square to both sides of the equation...actually, I think the method I learned was the one that @johnweldon1993 seems to be showing...it is refreshing my memory.

mathslover · Answer

It depends on your understanding actually, if you think John's method is what you have studied, then follow it... 

Both methods are correct

johnweldon1993 · Answer

^Indeed ...I have used @mathslover method before as well ...both are correct :)

kewlgeek555 · Answer

Oh, okay, thanks! :D

So, let me factor the left hand side...

(x + 2)(x + 2)

And to simplify the right hand ride would be...1.

So (x + 2)(x + 2) = 1 or...$$(x+2)^2=1$$

johnweldon1993 · Answer

Correct @kewlgeek555

kewlgeek555 · Answer

Yay. \o/
Thank you both. ;)

I should read the actual lesson and not just the review. (/)v(\)

johnweldon1993 · Answer

Yeah that might help lol...and no problem!

kewlgeek555 · Answer

Also, I probably have a few more, but the most important is one that I want to check if I did it right, I can tag you guys in a new question. cx