I'am in my 30's who's willing to relearn math from ground up and i really felt HN is the right place to get my problem solved so i've decided to post a couple of mathematical problems which i'am having trouble with.
if log(a-b/4) = log sqrta+log sqrtb,show that (a+b)^2 = 20ab
if x^2+y^2=8xy show that 2log(x+y) = log5+log2+logx+logy
decompose the following into partial fractions
1) 3x-1/[(x+2)(1-x+x^2)]
2) 1/(x^3+1)
3) 2x^2-14x+8/[(x^2+3x-2)(x-3)]
<p><pre><code> 1) 3x-1/[(x+2)(1-x+x^2)]
2) 1/(x^3+1)
3) 2x^2-14x+8/[(x^2+3x-2)(x-3)]
</code></pre>
In 1 and 3 are there supposed to be extra brackets? It's really hard to tell exactly what these are given that you've used no formatting.<p>Secondly, why are you trying to solve these? Where did they come from?<p>Solving the first one, assuming the numerator is supposed to be (3x-1), then we proceed as follows.<p>When we add fractions like a/b + c/d the result is (ad+bc)/bd. In these questions we assume we have the result of such an addition and try to compute all the terms.<p>Here we have a denominator of (x+2)(1-x+x^2) so we assume b=x+2 and d=1-x+x^2. that means we have the numerator formula is:<p><pre><code> (3x-1) = a(1-x+x^2) + c(x+2)
</code></pre>
In this case a and c are potentially polynomials (in fact by looking at the degrees c must be a polynomial of degree 1 larger than a).<p>Now just try setting a and c to small polynomials, c being 1 degree larger, and compute what you get on the right hand side. Does it look like the left hand side? Probably not. What do you need to change? How can you make it better.<p>Jiggle it around a bit and see what happens.<p>Tell me, what if a=(x+1) and c=(x^2-2) : what does the RHS evaluate to? Why is it wrong? How can you change a and c to make it better.<p>With regards the second question, are you aware that x^3+1 factorizes? Find a value of x that makes x^3+1 equal zero. What does that tell you about factors of x^3+1?<p>That's enough for now. I've give you a lot of questions. Usually you don't answer them all, just ignoring all but one. I'll see what you do this time. Also, I have to go now - I'll be back tomorrow.
heh, come to think of it, so is the first one:<p>it is given that log((a-b)/4) = log sqrta + log sqrtb, thus
(a-b)/4 = sqrt(ab),
thus, ((a-b)/4)^2 = ab,
thus, (a-b)^2 = 16ab,
thus a^2 - 2ab + b^2 = 16ab. Adding 4ab to both sides,
a^2 + 2ab + b^2 = 20 ab, giving
(a+b)^2 = 20ab.
What have you tried? What do you know?<p>Rearrange 2.log(a)=3.log(b)+5 to find b in terms of a.<p>Answering that will give us some idea of your current status the better to answer your questions.
the second one is easy:<p>it is given that X^2+Y^2=8XY,
thus, X^2+Y^2+2XY=10XY (i.e, added 2XY to both sides)
thus, (X+Y)^2 = 10XY. Taking log of both sides:
2 log (X+Y) = log 10 + log X + log Y = log 5 + log 2 + log X + log Y