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Mathematics 9 Online
OpenStudy (iwanttogotostanford):

How to do this specific problem by completing the square, PLEASE help: Write the quadratic function in vertex form. y = x^2 + 8x + 18

OpenStudy (iwanttogotostanford):

@welshfella

OpenStudy (iwanttogotostanford):

what are the exact simplified steps to solving a completing the square?

OpenStudy (welshfella):

first convert the x^2 + 8x to the form ( x + a)^2 - b first divide the coefficient of x by 2 ( that is 8 /2) = 4 then you subtract (a/2) ^ 2

OpenStudy (welshfella):

so we have x^2 + 8x = (x + 4)^2 - 4^2

OpenStudy (welshfella):

if you expand you'll see that this is a true identity (x + 4^2 - 16 = x^2 + 8x + 16 - 16 = x^2 + 8x

OpenStudy (iwanttogotostanford):

oh ok- will these "same" steps work for solving the other completing the square problems as well?

OpenStudy (welshfella):

Yes except if the coefficient of x s-s greater than 1 for example 4x^2 + 14x - you would first take out the 4 = 4 (x^2 + (14/4) x) and complete the square on whats in the parentheses

OpenStudy (iwanttogotostanford):

oh ok, and then so for this problem so far we have x^2+8x= (x+4)^2-4^2, so what do i do next?

OpenStudy (welshfella):

we have y = (x + 4)^2 - 16 + 18 y = (x+4)^2 + 2

OpenStudy (iwanttogotostanford):

how'd you get that

OpenStudy (welshfella):

you need to remember to divide the coefficient of x by 2 then subtract the square of this outside the brackets

OpenStudy (welshfella):

well it was originally y = x^2 + 8x + 18

OpenStudy (iwanttogotostanford):

oh ok got it

OpenStudy (mathmale):

You could check this result by multiplying out the square of (x+4) and adding 2. Does that take you back to the original y = x^2 + 8x + 18?

OpenStudy (welshfella):

if you can remember identities x^2 + ax = (x + (a/2)) ^2 - (a/2)^2

OpenStudy (iwanttogotostanford):

is that the identity of completing the square

OpenStudy (welshfella):

yes

OpenStudy (welshfella):

in our case a = 8 and of course a/2 = 4

OpenStudy (iwanttogotostanford):

oh ok, and are there any shortcuts to this?

OpenStudy (welshfella):

hmm. none that i know of!

OpenStudy (welshfella):

y = (x + 4)^2 + 2 is vertex form the graph would be a parabola with vertex (4 , 2)

OpenStudy (iwanttogotostanford):

ok!

OpenStudy (mathmale):

While the "formula" discussed above is completely correct, it might be easier to remember the following steps: y = x^2 + 8x + 18 1. Take half of the coefficient of x (which here is 8); 2. Square that result (which here produces 4^2 = 16) 3. Add that last result (16) to x^2 + 18x, and then subtract that same quantity: y = x^2 + 8x + 16 - 16 + 18, which simplifies to y=(x+4)^2+2 That's it. This last result has the form y=(x-h)^2+k, where (h,k) is the vertex. Can you now identify h and k? Write out the original equation, y = x^2 + 8x + 18, in this new format.

OpenStudy (welshfella):

Oh yes I see the error its x-h so the vertex will be at (-4,2)

OpenStudy (mathmale):

That's right. @iwanttogotostanford : Any further questions?

OpenStudy (iwanttogotostanford):

oh ok, thanks both of you helped a lot!

OpenStudy (welshfella):

yw

OpenStudy (iwanttogotostanford):

so @welshfella the answer that we concluded to is still right, correct?

OpenStudy (mathmale):

y=(x+4)^2 + 2 is indeed correct. Make certain that you know how to obtain the coordinates of the vertex. What are they?

OpenStudy (welshfella):

yes y = (x + 4)^2 + 2 is the function in vertex form The general form for the vertex is y = (x - h)^2 + k so comparing this with our function -h = 4 so h = -4 and k = 2 so the vertex (h , k) = (-4,2) They didn't ask for the coordinates of the vertex I just added it as an extra.

OpenStudy (welshfella):

its worth committing the formula for vertex to memory In UK we used to use a and b for the coordinates but now they use h and k . ( I don't know why!!)

OpenStudy (welshfella):

as practice can you complete the square of x^2 - 10x ?

OpenStudy (iwanttogotostanford):

yes I will try x^2+25 @welshfella

OpenStudy (iwanttogotostanford):

thats as far as i got... i followed the steps and... yeah..

OpenStudy (iwanttogotostanford):

did i do it right?

OpenStudy (mathmale):

Given: x^2 - 10x 1. find half of the coefficient of the x term. 2. Square this result. 3. Add this square, and then subtract this square, to x^2-10x. 4. Rewrite the perfect square trinomial as the square of a binomial. Refer to previous explanations; this is nothing new or different. Please show your work.

OpenStudy (welshfella):

gtg right now

OpenStudy (iwanttogotostanford):

@mathmale how do I do step 4 again? I did everything and got x^2-1x+25-25, but I don't know how to make that into a binomial

OpenStudy (iwanttogotostanford):

@mathmale

OpenStudy (mathmale):

1. find half of the coefficient of the x term. 2. Square this result. 3. Add this square, and then subtract this square, to x^2-10x. 4. Rewrite the perfect square trinomial as the square of a binomial. x^2 - 10x 1. half the coeff of the x term is half of -10, or -5. 2. The square of this result (-5) is +25. 3. Starting with x^2 - 10x, add 25, and then subtract 25: x^2 - 10x + 25 - 25 4. Rewrite x^2 - 10x + 25 as the square of a binomial

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