Another calculus question

Sully

What's the best way of breaking down below, so the zeros can be determined using the quadratic formula?

x^3+x^2-6x

Divide across by x leaving;
x^2+x-6 ?

See, the equation is actually [latex]\displaystyle\mbox{x^3 + x^2 - 6x = 0}[/latex], so if you divide one side by [latex]\displaystyle\mbox{x}[/latex], you have do divide the other too. The way you're doing it is, in some ways, correct; it just makes more logical sense, and it follows intuition, to do it the following way.

Factor out the [latex]\displaystyle\mbox{x}[/latex] on the left hand side, like this:

[latex]\displaystyle\mbox{x(x^2 + x - 6) = 0}[/latex], so now you know that one solution is [latex]\displaystyle\mbox{x = 0}[/latex] (because you have two factors, one is [latex]\displaystyle\mbox{x}[/latex], the other is the quadratic; if their product equals 0, then one of them must equal 0, if you get me), you can now use the quadratic formula on [latex]\displaystyle\mbox{x^2 + x - 6 = 0}[/latex] to find the other two solutions.

The way you did it (divide across by x) works alright, but it's always good to know this technique, because there are occasions where the division by x isn't possible.

Sully

So I can use that new equation for each calculation - zeros, max & min, inflection?

Edit: My approach correct here;
-x^3+x^2+x - -x(x^2+x+1)=0

Root x=0 and function x^2+x+1 ?

Sully

Sully wrote: »

So I can use that new equation for each calculation - zeros, max & min, inflection?

You can only use that to find the zeros. You'll have to use the original to find max. & min. and the points of inflection. (This is always the case).

Edit: My approach correct here;
-x^3+x^2+x - -x(x^2+x+1)=0

Root x=0 and function x^2+x+1 ?

I always take out a positive x, then leave the signs as they are inside the brackets. You see, if you multiply in that -x, you won't get the original equation. I'd give two tips, always make sure the first term of the equation is positive (which is typically the highest power of x, in this case the x cubed term), so if it's negative just multiply everything by -1 (i.e. just change everything's sign). The next tip I'd give would be to just take out a positive x, and leave the signs as they are inside the bracket, that way you won't encounter any mistakes with the signs.

Using your example:

[latex]\displaystyle\mbox{- x^3 + x^2 + x = 0}[/latex], multiply that by -1 (because the first term -- which is usually the highest power of x -- is negative. This gives you [latex]\displaystyle\mbox{x^3 - x^2 - x = 0}[/latex]. Now you can factor out the x (keep it positive) as you've done. This way there are no issues with the signs inside the brackets.

(P.S. I hope I haven't misunderstood your question somewhere).

Sully

So first step is changing the signs, then factorise the x? So in the above case..

x/x^3-x^2-x

x(x^2-x-1)=0 ?

Ok, just say your equation is [latex]\displaystyle\mbox{- x^3 + x^2 - 4x = 0}[/latex], the first thing that you want to do is to have the first term positive (the first term should also always be the highest power of x, so if the x^2 is before the x^3, rearrange it so it's in a descending order, if you get me).

You make the first term of the above equation postive, you just multiply everything by -1, giving [latex]\displaystyle\mbox{x^3 - x^2 + 4x = 0}[/latex]. Now, you factor out an x, giving [latex]\displaystyle\mbox{x(x^2 - x + 4) = 0}[/latex]. You now know that [latex]\displaystyle\mbox{x = 0}[/latex] is one solution, and you can solve the quadratic with the quadratic formula.

You have to use the original in any calculus operations (such as finding the max. and min. or finding the points of inflection). The above only works for finding the zeros (or roots) of the equation. Also, your original division method works in these examples just fine, but it's good to know this technnique incase a problem arises where the division doesn't work so easily.

P.S. It's no necessary to change the sign of the first term and have the powers in descending order, but it makes it a lot easier, and it's harder to make a mistake, if you follow the above.

P.P.S. This won't work if the equation contains a constant, like this one for example: [latex]\displaystyle\mbox{x^3 + x^2 - 4x + 7 = 0}[/latex]. If the equation contains a constant (7, in this case), you'll have to use a method such as the factor theorem to find the first root.

Sully

Ok here is an example I did;

[latex]-x^3%2Bx^2%2Bx[/latex]

Becomes..

[latex]x^3 - x^2 -x[/latex]

Becomes..

[latex]x^2-x-1[/latex]

Quadratic formula work out and just as we work everthing out, we get a square root of 5 giving a decimal point rather then a whole number. Is that normal?

Yah, that's exactly right. Your first solution is x = 0, then the two others contain [latex]\displaystyle\sqrt{5}[/latex].

If you go here, it'll give you the three solutions (just work them out in decimal form and compare them to your own solutions). It's a pretty handy website for checking whether your solutions are correct.

Sully

It seems to have a half before my square root - why is that? I was left with

1+ [latex]\displaystyle\sqrt{5}[/latex] = 1 = 2.236 = 3.236
and
1- [latex]\displaystyle\sqrt{5}[/latex] = -1.236

Also, your tip about removing the first term negative so its a positive - can that be used all the time for example; when using the quadratic formula for max/min etc. You mentioned I had to use the original equation, which was a cubed - I assume I just differentiate the two as normal, put it equal to zero and solve?

henbane

[latex]\frac{1 \pm \sqrt{(-1)^2 -4(1)(-1)}}{2(1)}[/latex]

The half comes from the 2a

Sully

henbane wrote: »

[latex]\frac{1 \pm \sqrt{(-1)^2 -4(1)(-1)}}{2(1)}[/latex]

The half comes from the 2a

Ya that's perfect, forgot about 2a

Sully

henbane wrote: »

[latex]\frac{1 \pm \sqrt{(-1)^2 -4(1)(-1)}}{2(1)}[/latex]

The half comes from the 2a

Why is it not the way I suggested? Is there some rule I am missing which converts the 2a into a half times the usual method?

[latex]\frac{-1\pm\sqrt{(5)}}{2(1)}[/latex]

henbane

Sully wrote: »

Why is it not the way I suggested? Is there some rule I am missing which converts the 2a into a half times the usual method?

I'm not sure what you mean but
[latex]\frac{1}{2} (1 \pm \sqrt{5}) = \frac{1 \pm \sqrt{5}}{2}[/latex]

Sully

henbane wrote: »

I'm not sure what you mean but
[latex]\frac{1}{2} (1 \pm \sqrt{5}) = \frac{1 \pm \sqrt{5}}{2}[/latex]

Is that a rule? I came across it in another question also. My way of calculating it gave different results to the suggested method here?

henbane

It's just algebra, e.g. [latex]\frac{3}{2} = \frac{1}{2} * 3 = \frac{1+2}{2} = \frac{1}{2}(1+2)[/latex]
Or do you mean the rule for solving quadratics is different from the one you're using?

Sully

It was the algebra that was confusing me. Cant say I have seen that (the second half of your expression) in a while!

henbane

Actually algebra is probably the wrong word but I'm never very good on terminology. I reckon it's mostly used as a neater way of writing the result as the full fraction takes up two lines. It's just a matter of getting used to the representations.

Sully

henbane wrote: »

Actually algebra is probably the wrong word but I'm never very good on terminology. I reckon it's mostly used as a neater way of writing the result as the full fraction takes up two lines. It's just a matter of getting used to the representations.

Id have though keeping things in decimals was fine if you didnt want to use fractions. At least, thats the way I was thought. But it seems to give different answers!

henbane

You lose precision with decimals so it's generally best to leave any decimal representation until the end of your calculations