Simple Question Regarding Quadratic Equations (pre-calc)

[TalkOrigin]

Hi, I've been attempting to teach myself mathematics in preparation for further study and I'm currently going through quadratic equations. In case it helps, I'm studying to go back to university, I'll be doing a physics degree with a foundation year. I'm just trying to learn as much math as possible before I start, and I'm working through the UK A-Level Math syllabus, OCR, I'm on C1 at the moment. I seem to be fine with quadratic equations except for when the equation is in the form $c + bx - a x^2$ instead of $a x^2 + bx + c$

I know if it's $c + bx + a x^2$ I can just flip everything round so it's in the standard form and it doesn't give me too much trouble, but when the coefficient of $x^2$ is a negative number, I just don't know how to approach the problem. Here is one such problem I'm stuck on:

"Use the completed square form to find as appropriate the least or greatest value of each of the following expressions, and the value of x for which this occurs".

$4 + 6x - x^2$

I'm sure it's simple, I just don't know how to do it. I've tried looking online but I can't find anything addressing how to solve a question like this specifically. I'm probably just being an idiot and missing some incredibly obvious step. My attempt at solving it was:

$4 + 6x - x^2$

$4 + (3-x)^2 - 9$

then, do $4 + -9 = -5$ So

$(3-x)^2 - 5$

Least Value - I got -5, although apparently the answer is 13. I know I can't be a million miles off as I don't think it's coincidence that 4+9=13, I just can't see where I'm going wrong. I got the value of x for when this least value occurs correct somehow, at 3.

Anyway thanks for the help, I don't know why I get so confused when the equations are formulated in this way. I don't even need someone to tell me what the answer is, but how to solve an equation when it's in that form would be great, maybe to point out where I'm going wrong also.

Thanks again

 

[eumyang]

First off, a minor nitpick: you don't have an equation, just an expression, because there is no equal sign.

You have to rearrange the quadratic in descending powers of x. What you are doing is rewriting the equation in vertex form:
$y = a(x - h)^2 + k$,
What you've written is not in vertex form; the x must appear first inside the parentheses. If there is a negative in front of the x², factor out the negative:
$-x^2 + 6x + 4 = 0$
$-(x^2 - 6x) + 4 = 0$

Now try to complete the square.

 

[TalkOrigin]

First off, a minor nitpick: you don't have an equation, just an expression, because there is no equal sign.

You have to rearrange the quadratic in descending powers of x. What you are doing is rewriting the equation in vertex form:
$y = a(x - h)^2 + k$,
What you've written is not in vertex form; the x must appear first inside the parentheses. If there is a negative in front of the x², factor out the negative:
$-x^2 + 6x + 4 = 0$
$-(x^2 - 6x) + 4 = 0$

Now try to complete the square.

Firstly, thank you so much for the quick response. About it not being an equation, that's my bad, like I said I'm just getting into maths again for the first time in over a decade so I'm a bit sloppy when it comes to language. Just for my own interest though, whenever you see a quadratic expression like this, can it be assumed that the expression = y ?

What you wrote makes perfect sense, I just need to factor out the negative first. I wasn't aware the x must appear first inside the parenthesis either.

Can I show you my working and you can let me know if I went about this the right way now? I'm not sure about one step, right near the end.

$4 + 6x - x^2$

Put it in standard form:

$-x^2 + 6x +4$

Factor out negative:

$- ( x^2 - 6x ) + 4$

Now complete the square:

$- [ ( x -3 )^2 - 9 ] + 4$

I know that next step is to take $- 9$ out of the negative brackets, so:

$- ( x - 3 )^2 + 9 + 4$

So now I get the 13, because now we have

$- ( x - 3 )^2 + 13$

I'm just not sure what we do with the negative in front of the parenthesis, is this the completed square? I understand now that as long as whatever is in the parenthesis = 0, it doesn't matter what's directly before it. So I think (hope) I get it!

Again, thanks for the help, I know I'm not a genius when it comes to this stuff so I appreciate that this is probably somewhat laborious for you.

 

[eumyang]

Firstly, thank you so much for the quick response. About it not being an equation, that's my bad, like I said I'm just getting into maths again for the first time in over a decade so I'm a bit sloppy when it comes to language. Just for my own interest though, whenever you see a quadratic expression like this, can it be assumed that the expression = y ?

This is a quadratic expression: $ax^2 + bx + c$
This is a quadratic equation: $ax^2 + bx + c = 0$
This is a quadratic function: $f(x) = ax^2 + bx + c$ or $y = ax^2 + bx + c$
(Note that a cannot be 0 in any of the above.)

$- ( x - 3 )^2 + 13$

I'm just not sure what we do with the negative in front of the parenthesis, is this the completed square?

It is the completed square. The negative indicates that 13 is the greatest value, not the least value.
If you have the expression in the completed square form like this:
$a(x - h)^2 + k$,
and a is positive, then k is the least value. If a is negative, then k is the greatest value.

 

[TalkOrigin]

This is a quadratic expression: $ax^2 + bx + c$
This is a quadratic equation: $ax^2 + bx + c = 0$
This is a quadratic function: $f(x) = ax^2 + bx + c$ or $y = ax^2 + bx + c$
(Note that a cannot be 0 in any of the above.)


It is the completed square. The negative indicates that 13 is the greatest value, not the least value.
If you have the expression in the completed square form like this:
$a(x - h)^2 + k$,
and a is positive, then k is the least value. If a is negative, then k is the greatest value.

Thanks, that's clear now. I just used the method you outlined on another question and I'm now able to solve these types of questions... Little victories lol.