Quadratic expression in completed-square form help

[matheus]

Hey guys,

I was wondering if I could get a little bit of help here with these questions regarding quadratic expressions and the completed-square form I’m being asked to write it in.

(i) Write the quadratic expression x² -32x + 14 in completed-square form.

(ii) Use the completed square form from part (i) to solve the equation x² - 32x + 14 = 0, leaving your answer in exact (surd) form.

(iii) Use the completed square form from part (i) to write down the vertex of the parabola y = x² - 32x + 14.

I just can’t seem to get my head around it and I really would like to understand it. Youtube has helped a bit but I’m having trouble applying that with these specific questions below.

I will keep trying in the meantime; think I need a coffee or a red bull ha ha

Thank you in advance if anyone is able to help here!

Have a great day

 

[stapel]

(i) Write the quadratic expression x² -32x + 14 in completed-square form.

(ii) Use the completed square form from part (i) to solve the equation x² - 32x + 14 = 0, leaving your answer in exact (surd) form.

(iii) Use the completed square form from part (i) to write down the vertex of the parabola y = x² - 32x + 14.

I just can’t seem to get my head around it

The process is fairly straight-forward. Where are you getting stuck?

For instance, in this case, you grouped the variable-containing terms together, leaving space for what you'll be adding (or subtracting):

. . . . .\(\displaystyle x^2\, -\, 32x\, \) _________ [/tex]\, +\, 14[/tex] _________

Then you divided the coefficient of the linear term in half, keeping track of the sign:

. . . . .\(\displaystyle \dfrac{1}{2}\left(-32x\right)\, =\, -16\)

Then you squared this:

. . . . .\(\displaystyle (-16)^2\, =\, 256\)

Then you added this to the quadratic (to "complete" the squared binomial that you're creating) and subtracted this from the constant term (so that, net, you're adding zero, so you haven't changed anything). Then, using the sign you remembered to note, you completed the square, converted to binomial form, simplified the constant terms, and... then what?

Please be complete, so we can see where you're having trouble. Thank you!

 

[Ishuda]

Hey guys,

I was wondering if I could get a little bit of help here with these questions regarding quadratic expressions and the completed-square form I’m being asked to write it in.

I just can’tseem to get my head around it and I really would like to understand it. Youtube has helped a bit but I’m having trouble applying that with these specific questions below.

I will keep trying in the meantime; think I need a coffee or a red bull ha ha

Thank you in advance if anyone is able to help here!

Have a great day

The completed square form is based on the equation
(x+a)² = x² + 2 a x + a²

To apply this let's start with an expression of the form Ax²+Bx+C where A is not zero. The first thing we notice is that our completed square form template above has a 1 for the coefficient of x and this expression has an A. Well just factor out an A to get A[x²+2(B/(2A))x+(C/A)]. Next, we note that we have a 2 as part of the coefficient for x in our form above. We can get that by multiplying and dividing the x coefficient of our expression by 2 to get A[x²+2(B/(2A))x+(C/A)]. Now we know what a is in the form above, a=B/(2A). The next step is to add and subtract a² = (B/(2A))² to get
A [x² + 2 (B/(2A)) x + (B/(2A))² + (C/A)-(B/(2A))²]
which, using our form above, we can write as
A { [x + (B/(2A))]² + (C/A) - (B/(2A))² }

Now, lets do an example with the expression x² - 12x + 35. The A is equal to 1, so we don't have to do anything about that. The next step is to put in the 2 for the x coefficient. So divide 12 by 2 to get
x² - 12x + 35 = x² - 2(12/2)x + 35 = x² - 2*6 x + 37 = x² + 2*(-6) x + 35.
Now add and subtract a (= -6) squared to get
x² - 12x + 35 = x²+2*(-6)x+(-6)²+35-(-6)² = [x²+2*(-6)x+(-6)²]+35-36 = [x²+2*(-6)x+(-6)²] - 1.
And finally, using our form above,
x² - 12x + 35 = [x²+2*(-6)x+(-6)²] - 1 = [x+(-6)]² + 1 = (x-6)² - 1

If you show us your work on your problem, we can see where you are having difficulties and possibly help.

 

[matheus]

Hey guys,


Thanks so much for taking the time to reply and help me out here, massively appreciate it!


I didn’t realise I had some replies so sorry for the slow return.


I have been working on these questions and have them nearly finished (hopefully) so if it would be ok I will post up my work and politely ask if you can spot any silly mistakes.


Thank you!

 

[Ishuda]

Hey guys,


Thanks so much for taking the time to reply and help me out here, massively appreciate it!


I didn’t realise I had some replies so sorry for the slow return.


I have been working on these questions and have them nearly finished (hopefully) so if it would be ok I will post up my work and politely ask if you can spot any silly mistakes.


Thank you!

Sure - go ahead and post with your work. It would probably be better if there were one problem per post.

 

[matheus]

Hey guys,

Thanks again for the replies here, excellent!

So here is my working so far, massively grateful of any error spotters, ha ha!

Many thanks!