Please help! Algebra question!

skyscraperblue

Two algebra questions that we got for revision homework. I'm decent at maths and thought they looked easy until I tried them. First one...

Given the equation ax^2 + bx + c = 0 (a,b,c E R)
If 1 - (sqrt)2 is one of the roots of the equation write down the other root. Find the values of a, b and c.

The other one I thought would be grand, divide in the factor, sure, until none of my division gave me anything close to the 'show' part...

(x - t)^2 is a factor of 2x^3 - tx^2 + hx + k, h and k E R
Show that 9h^3 + 64k^2 = 0.
Find in terms of t all three roots of the equation 2x^3 - tx^2 + hx + k = 0.

Please help! Anyone got any ideas?

click_here!!!

For the first one, it might be the conjugate. I.e. the other root might be 1 + sqrt(2).
a = 1
b = minus the sum of the two roots.
c = the product of the roots.

For the second one, you use the Factor Theorem. f((x+t)^2) = 0. I think it was on the Leaving Cert Maths course.

skyscraperblue

click_here!!! wrote: »

For the first one, it might be the conjugate. I.e. the other root might be 1 + sqrt(2).
a = 1
b = minus the sum of the two roots.
c = the product of the roots.

For the second one, it looks like you use the Factor Theorem.
Divide 2x^3 - tx^2 + hx + k by (x - t)^2. Then if the answer is in the form (x +/- something), then -/+ something = 0. (note the opposite sign)

Hope this helps.:)

Thanks! I tried the Factor Theorem a ton of times though, because I thought of it straight away too. When I divide, my remainder line is (h+4t^2)x + k - 3t^3. Which doesn't really give me anything I can see any way of working with because the t's are stuck in there.

bnt

For the first one, I agree that the other would be the conjugate 1+sqrt(2), since that's the form of result produced by the Quadratic Equation:
[latex]\displaystyle
{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}[/latex]

You can use that equation and the info you have to generate a couple of simultaneous equations, but you need more than that. Don't make the mistake of inferring that b = -1 from that alone: instead, [latex]\frac{-b}{2a}=1[/latex]. Strictly speaking, you can't get the value of a from the roots alone, since you can draw an infinite number of parabolas through any two roots. So if you're going to say a=1, that's an assumption. Either that or we're missing some information.

red_fox

bnt wrote: »

Strictly speaking, you can't get the value of a from the roots alone, since you can draw an infinite number of parabolas through any two roots. So if you're going to say a=1, that's an assumption. Either that or we're missing some information.

That's a good point and of the type often missed even by those setting the question. My guess would be that a=1 is assumed but it should ask to find b and c such that those are the roots of [latex]x^2+bx+c=0[/latex]. Allowing a to be any real number means you have a set of solutions [latex]\{(\lambda,\lambda b,\lambda c) \mid \lambda \in \mathbb{R} \setminus \{0\} \}[/latex].

And for some reason \mathbb isn't giving me what I want.

Michael Collins

Picking the question apart some more: since the coefficients of the original quadratic are allowed to be real, it isn't actually justifyable to say that one of the roots is going to be the radical conjugate of the other.

If the coefficients were rational, then it would necessarily be the case.

Not a well stated question really.

MathsManiac

Agreed. Much nicer, if one didn't want to assume the sum and product were going to be integral, would be:
...where a,b,c E Z, and gcd(a,b,c)=1.

MathsManiac

Anyway, on the second problem, sometimes the factor theorem won't do the trick if you have a double root, as you end up one equation short.

Instead, you could divide the quadratic factor into the cubic and set the remainder equal to 0, (that is 0x+0).

Alternatively, you could use the fact that if t is a double root of f(x), then it is a root of f'(x), (the derivative of f).

skyscraperblue

MathsManiac wrote: »

Instead, you could divide the quadratic factor into the cubic and set the remainder equal to 0, (that is 0x+0).

This was my first instinct and exactly what I tried to do at least five times over. But it still won't give me what I want - anything I end up in has t's in it that I can't get rid of.

Michael Collins

red_fox wrote: »

...And for some reason \mathbb isn't giving me what I want.

Yeh it doesn't seem to be included in the MathTran implementation of TeX.

This is the closest I can get:

\rm I\!R gives [latex] \rm I\!R[/latex]

and

\rm I\!N gives [latex] \rm I\!N[/latex]

or you could just go with boldface to denote sets, which is the usually the prefered option anyway when it is possible - Blackboard typeface was intented for, err, blackboards or handwritten maths, when you couldn't easily make your text boldface.

\bf R gives [latex] \bf R[/latex]

or

\bf N gives [latex] \bf N[/latex]

MathsManiac

skyscraperblue wrote: »

This was my first instinct and exactly what I tried to do at least five times over. But it still won't give me what I want - anything I end up in has t's in it that I can't get rid of.

You should have got h + 4t^2 = 0 and k - 3t^3 = 0.

If you use each of those separately to write t^6 and equate, you're there.

Finding the roots is easy, since you've been given two factors, and you've already worked out the other factor using your diveision.