Completing the square: -x^2 + 30x + 125, etc.

[Monkeyseat]

A few here:

1) -x^2 + 30x + 125

Goes to:

-1(x-15)^2 - 100

Or is it -1(x+15)^2 - 100?

In regard to Q1, after going on AlgebraHelp.com I realised it could be -1(x-15)^2 + 225 + 125 which will simplify to -1(x-15)^2 + 350 ?

This completed square shows the area of a zone, how do I find the maximum area from this? According to the answers it's 350 units but I don't know how to get this. I think it's when x=0 but I don't understand.

2) f(x) = 3x^2 + 12x + 17

Goes to:

f(x) = 3(x+2)^2 + 5

How do I "deduce the greatest value of 1/f(x)?"

3) y = 4x - x^2

Goes to:

y= -1(x-2)^2 + 4

Or is it -1(x+2)^2 + 4?

Many thanks for any help.

 

[stapel]

1) -x^2 + 30x + 125

Goes to: -1(x-15)^2 - 100

What were your steps between the original form and the completed-square form? How did you get the "100" term?

Same for the others....

Please be complete. Thank you!

Eliz.

 

[Monkeyseat]

1)

Okay, it's a bit hard to explain this one. There was a diagram with various dimensions and when multiplied together it came like this:

(x + 5)(-x + 25) + 10x

= -x^2 + 25x - 5x + 125 + 10x
= -x^2 + 30x + 125

(x + 15)^2 = x^2 + 15x + 15x +225
= x^2 + 30x + 225

Therefore:

-1(x - 15)^2 - 225 + 125
= -1 (x - 15)^2 - 100

I think that might be wrong. After playing with the calculator on AlgebraHelp.com I realised it might be -1(x-15)^2 + 225 + 125 which then simplifies to -1(x-15)^2 + 350 ?

How do I find the max area? The book says the max value is 350 but I don't know why... I think it is when x=15 but I don't get it.

2)

f(x) = 3x^2 + 12x + 17
= 3(x^2 + 4x) + 17
= 3((x + 2)^2 - 4) + 17
= 3(x + 2)^2 + 5

I got 5 as the lowest value when x = -2 but how do I "deduce the greatest value of 1/(f)x?"

3)

y = 4x - x^2
= -x^2 + 4x + 0

(x + 2)^2 = x^2 - 2x - 2x +4
= x^2 - 4x + 4

When multiplied by negative -1 to get -x:

-1 (x - 2)^2 = -x^2 - 4x - 4

Therefore goes to: -1(x - 2)^2 + 4

I think this is wrong.

Might be a few typos.

EDIT:

Corrected most in both posts now. Sorry.

 

[stapel]

1) Okay, it's a bit hard to explain this one. There was a diagram with various dimensions and....

Hmm... If there are "diagrams" and such, then perhaps your book is making this too hard...? :shock:

Instead, try this standardized algorithm, explained with an example problem:

. . . . .Take the quadradic:

. . . . . . . .3x² - 12x + 7

. . . . .Factor out whatever is in front of the variable terms:

. . . . . . . .3(x² - 4x) + 7

. . . . .Find half of the linear coefficient, keeping track of
. . . . .the sign:

. . . . . . . .(1/2)(-4) = -2

. . . . .Square this, and add inside the parentheses.
. . . . .Balance this addition inside with subtraction
. . . . .outside, using whatever you factored out:

. . . . . . . .3(x² - 4x + 4) + 7 - (3)(4)

. . . . .Convert the variable part into squared form,
. . . . .using the value, with its sign, that you found
. . . . .above:

. . . . . . . .3(x - 2)² + 7 - (3)(4)

. . . . .Simplify:

. . . . . . . .3(x - 2)² + 7 - 12

. . . . . . . .3(x - 2)² - 5

As you can see, the steps can be neatly shown, and the reasoning clearly stated. :wink:

In your case, the steps would look something like this:

. . . . .-x² + 30x + 125

. . . . .-1(x² - 30x) + 125

. . . . .half the linear term: (1/2)(-30) = -15
. . . . .squared: (-15)² = 225

. . . . .-1(x² - 30x + 225) + 125 - (-1)(225)

. . . . .-1(x - 15)² + 125 + 225

...and so forth.

Eliz.

 

[pka]

You know that \(\displaystyle f(x) = 3x^2 + 12x + 17 = 3\left( {x + 2} \right)^2 + 5\).
You also know that \(\displaystyle 3\left( {x + 2} \right)^2 \ge 0\) therefore you know that \(\displaystyle f(x) \ge 5\) and \(\displaystyle f(x) = 5\mbox{ only if }x = - 2.\)
Thus \(\displaystyle f(x) \ge 5\quad \Rightarrow \quad \frac{1}{{f(x)}} \le \frac{1}{5}.\)

 

[Monkeyseat]

Because my post was moved I'll post my other question here, is -1(x - 2)^2 + 4 right for Q3?

Question 1 was moved into this thread: http://www.freemathhelp.com/forum/viewt ... 8130#98130

[Below is copied and pasted about Q2 from before my post was moved]

EDIT:

Sorry my internet went down so I didn't see your reply pka. I don't really understand what you're saying. I'm only doing A Level maths.

EDIT EDIT:

Actually after reading it again I kind of understand you now pka. So although x could be bigger than 5 with 6 and make the completed square/factorisation bigger (?), it couldn't be because if x was bigger than 5 i.e. 6 the fraction 1/f(x) would be smaller i.e. 1/6? You couldn't have 1/4 because 1/5 is the smallest fraction you can get due to the +5 at the end? It's best to make it as small as possible by making the brackets 0? Is 5 the only value possible that will ever work (not necessarily biggest)? Thanks! Sorry for asking all those questions. I can clear up what's confusing me a bit if needs be. I'm just not sure what you meant when you said:

\(\displaystyle \Rightarrow \quad \frac{1}{{f(x)}} \le \frac{1}{5}.\)

I thought it was the other way around, I don't know what the arrow signifies. Please can you explain this step? Thank you.

 

[galactus]

Because my post was moved I'll post my other question here, is -1(x - 2)^2 + 4 right for Q3?

To see if it is, just expand it back out. Do you get the original?.

I think you do.

Just as an aside, do not write the 1 in front.
There's no need.
\(\displaystyle \L\\-(x-2)^{2}+4\)

When completing the square, it may help to remember the formula:

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}\)

Just plug in your values. a=-1, b=4, c=0

 

[Monkeyseat]

Because my post was moved I'll post my other question here, is -1(x - 2)^2 + 4 right for Q3?

To see if it is, just expand it back out. Do you get the original?.

I think you do.

Just as an aside, do not write the 1 in front.
There's no need.
\(\displaystyle \L\\-(x-2)^{2}+4\)

When completing the square, it may help to remember the formula:

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}\)

Just plug in your values. a=-1, b=4, c=0

I'm having trouble multiplying out when it's got a negative number in front. Do I just multiply out then invert the signs? That's why I didn't know if it was (x+...) or (x-...). I thought about leaving the 1 bit off with just the negative sign (-) like you said.

When completing the square for -x^2 + 30x + 125 I changed the sign from -(x + 15)^2 + 100 I think to -(x - 15)^2 + 350 so I am confused now as to why. When completing the square for 4x - x^2 I didn't change the sign and left it as -1(x - 2)^2 + 4. I don't know which it is. Why do you change the sign? I only did it for one and I've forgotten now. :lol:

Thanks.

EDIT:

Corrected negative sign sorry.

 

[galactus]

No need for a negative for this one:

\(\displaystyle \L\\x^{2}+30x+125=0\)

\(\displaystyle \L\\x^{2}+30x=-125\)

The square of half the coefficient of x:

\(\displaystyle \L\\x^{2}+30x+225=-125+225\)

The left is a perfect square, factor:

\(\displaystyle \L\\(x^{2}+15)^{2}=100\)

\(\displaystyle \L\\(x^{2}+15)^{2}-100=0\)

Or, just plug into the formula and get the same.

Just for kicks, let's derive the formula for the general case.

\(\displaystyle \L\\ax^{2}+bx+c=0\)

\(\displaystyle \L\\ax^{2}+bx=-c\)

Divide by a:

\(\displaystyle \L\\x^{2}+\frac{b}{a}x=\frac{-c}{a}\)

The square of half the coefficient added to both sides:

\(\displaystyle \L\\x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}=\frac{-c}{a}+\frac{b^{2}}{4a^{2}}\)

The left is a perfect square, so factor:

\(\displaystyle \left(x+\frac{b}{2a}\right)^{2}=\frac{-c}{a}+\frac{b^{2}}{4a^{2}}\)

On the right, factor out 1/a:

\(\displaystyle \L\\\left(x+\frac{b}{2a}\right)^{2}=\frac{1}{a}\left(-c+\frac{b^{2}}{4a}\right)\)

Multiply by a:

\(\displaystyle \L\\a\left(x+\frac{b}{2a}\right)^{2}=-c+\frac{b^{2}}{4a}\)

\(\displaystyle \L\\\fbox{a\left(x+\frac{b}{2a}\right)^{2}+c-\frac{b^{2}}{4a}}\)

There's the formula that'll work every time.

I used the same steps as in any completing the square problem, only in general.

 

[Deleted member 4993]

No need for a negative for this one:

\(\displaystyle \L\\x^{2}+30x+125=0\)

\(\displaystyle \L\\x^{2}+30x=-125\)

The square of half the coefficient of x:

\(\displaystyle \L\\x^{2}+30x+225=-125+225\)

The left is a perfect square, factor:

\(\displaystyle \L\\(x+15)^{2}=100\)...............small typo fixed

\(\displaystyle \L\\(x+15)^{2}-100=0\)

 

[Monkeyseat]

I believe you've made another typo galactus (although it was my fault because I left the negative sign out in that reply, but it was in on my original topic first post):

-x^2 + 30x + 125

It's negative x, you did not do negative x I believe from looking at it. Please can you show me with the negative sign?

I'd appreciate it if you read my above post because I don't know how/why/when to change the sign in the brackets, like I stated in the example in that post:

Because my post was moved I'll post my other question here, is -1(x - 2)^2 + 4 right for Q3?

To see if it is, just expand it back out. Do you get the original?.

I think you do.

Just as an aside, do not write the 1 in front.
There's no need.
\(\displaystyle \L\\-(x-2)^{2}+4\)

When completing the square, it may help to remember the formula:

\(\displaystyle \L\\a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a}\)

Just plug in your values. a=-1, b=4, c=0

I'm having trouble multiplying out when it's got a negative number in front. Do I just multiply out then invert the signs? That's why I didn't know if it was (x+...) or (x-...). I thought about leaving the 1 bit off with just the negative sign (-) like you said.

When completing the square for x^2 + 30x + 125 I changed the sign from -(x + 15)^2 + 100 I think to -(x - 15)^2 + 350 so I am confused now as to why. When completing the square for 4x - x^2 I didn't change the sign and left it as -1(x - 2)^2 + 4. I don't know which it is. Why do you change the sign? I only did it for one and I've forgotten now. :lol:

Thanks.

Thanks.

If someone could explain what pka said about Q2:

Thus \(\displaystyle f(x) \ge 5\quad \Rightarrow \quad \frac{1}{{f(x)}} \le \frac{1}{5}.\)

That would also be great. I kind of understand it a bit but I'm not sure about why they are bigger/smaller than or equal to.

So:

1) Is Q1 correct? Is the max value 15 because anything else would make it smaller (the brackets would go negative)?

2) Can someone clear the f(x) thingy up in Q2 that pka mentioned (see above posts)?

3) Is Q3 right?

 

[pka]

You are just being blind to the obvious.
If \(\displaystyle a \ge 5\) this is it not obvious that \(\displaystyle \frac{1}{a} \le \frac{1}{5}\)?
That is all there is to it!
Thus the maximum for \(\displaystyle \frac{1}{{f(x)}}\) is \(\displaystyle \frac{1}{5}\).

 

[Monkeyseat]

Sorry. I understand now.

Thanks.