Question

Solve the equation by completing the square. x^2+12x+18=0. Please explain what completing the square means.

Answer 1

do you know how to solve using quadractic formulae (not that this is how you should do this - I just want to know if you are aware of it)?

Answer 2

yes

Answer 3

ok, so the quadratic formulae can be derived by whats called "completing the square"

Answer 4

in general, we have:\[ax^2+bx+c=0\]we first divide by 'a' to get:\[x^2+\frac{bx}{a}+\frac{c}{a}=0\]

Answer 5

the next step is to just concentrate on the x^2 and x terms - forget the constant

Answer 6

and we see if this can be written in the form (x+q)^2

Answer 7

this is what "completing the square" is about

Answer 8

how?

Answer 9

let me show you....

Answer 10

6

Answer 11

36

Answer 12

\[(x+\frac{b}{2a})^2=x^2+\frac{bx}{a}+\frac{b^2}{4a^2}\]

Answer 13

@tapdancer8: do you understand the expansion I just wrote?

Answer 14

6^2 = 36 good, now add 36 to the right side like this x^2+12x + 36 +18
Now can you factor the x^2+12x + 36 part?

Answer 15

@asnaseer: sort of, it's an expanded version of the quadratic formula

Answer 16

we had this earlier:\[x^2+\frac{bx}{a}+\frac{c}{a}=0\]and I just showed you that:\[(x+\frac{b}{2a})^2=x^2+\frac{bx}{a}+\frac{b^2}{4a^2}\]so we can combine the two next

Answer 17

@moses_:yes, it becomes (x+6)^2

Answer 18

\[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}=x^2+\frac{bx}{a}\]so put this into our 1st equation to get:\[(x+\frac{b}{2a})^2-\frac{b^2}{4a^2}+\frac{c}{a}=0\]

Answer 19

then rewrite as:\[(x+\frac{b}{2a})^2=\frac{b^2}{4a^2}-\frac{c}{a}\]

Answer 20

this is now in the form of:
(something)^2 = constant

and your next step would be:
something = squareroot(constant)

Answer 21

good, thats essentially what asnaseer is trying to show you, but you can see it faster this way. The whole point is so you can now factor a part of the left hand side of the equation. Now you also must subtract 36 on the left hand side of the equation so it comes out to be the exact same as when we started so you have x^2+12x + 36 +18-36. No you already factored the left hand side first 3 terms and as you can see by subtracting the 36 you have not really done anything illegal in terms of the rules of math. so you turned x^2+12x + 36 into (x+6)^2 and you are left with 18-36 so what is that all combined? in the new form?

Answer 22

this is what completing the square is about.
so, in your case:\[x^2+12x+18=0\]you would first see that:\[(x+6)^2=x^2+12x+36\]and use this to get:\[(x+6)^2-18=0\]which leads to:\[(x+6)^2=18\]and solve from there

Answer 23

so completing the square essentially involves:
1) divide by the coefficient of x^2
2) take half the resulting coefficient of x to form the (x+p)^2 term

Answer 24

in your case the coefficient of x^2 was just 1 - so no need to do any initial division

Answer 25

\[\left( x+6 \right)^2=-18\]\[\sqrt{\left( x+6 \right)^2}=\sqrt{-18}\]\[x+6=\pm i \sqrt{18}\]\[x=-6\pm 3i \sqrt{2}\]

Answer 26

right?

Answer 27

you have the sign wrong

Answer 28

which one?

Answer 29

\[(x+6)^2=18\]

Answer 30

the -18 is 18

Answer 31

but you just solved his equation...you should really take a look at how he got there.

Answer 32

how? if I had 18-36?

Answer 33

ok, you agree the following is true?\[(x+6)^2=x^2+12x+36\]

Answer 34

yes

Answer 35

good, so next we did this:\[(x+6)^2-18=x^2+12x+36-18=x^2+12x+18\]make sense so far?

Answer 36

yes

Answer 37

right, now your original equation is just the right-hand-side of what I just wrote

Answer 38

so your original:\[x^2+12x+18=0\]can be rewritten as:\[(x+6)^2-18=0\]do you see?

Answer 39

yeah

Answer 40

good, so next we just add 18 to both sides to get:\[(x+6)^2=18\]

Answer 41

oh

Answer 42

and you know how to solve from there

Answer 43

tap re-do the problem and you will see it. Dont just follow him along. you didnt follow how he got to that part...

Answer 44

I can try and explain it more simply by working through another example if you want?

Answer 45

yes please…do you want another from my homework or were you going to make it up?

Answer 46

either - which do you prefer?

Answer 47

homework. 7t^2+28t+56=0

Answer 48

ok, so if you recall the steps I listed above, these were:
1) divide by the coefficient of x^2 first

so here we have:\[7t^2+28t+56=0\]so 1st we divide by 7 to get:\[t^2+4t+8=0\]step 2) was to half the coefficient of the x-term to form the square, so here we get:\[(t+2)^2=t^2+4t+4\]but we want:\[t^2+4t+8\]so we need to add on another 4 to get:\[(t+2)^2+4=t^2+4t+8\]so we can now use this in the original to get:\[(t+2)^2+4=0\]giving:\[(t+2)^2=-4\]

Answer 49

do the steps make sense?

Answer 50

yes. thanks so much.

Answer 51

no problem - I'm glad we got there in the end :-)

Answer 52

the basic step involved here is, when you have something like:\[x^2+bx+c=0\]to recognise that:\[(x+p)^2=x^2+2px+p^2\]and use this

Answer 53

here 2p must equal b -- which is why we have to half the coefficient of the x-term when forming the square

Answer 54

ok. have to go to dance. thanks again!!

Answer 55

ok - enjoy your dancing...