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Two issues...

by mixpanda » Sat May 02, 2009 11:22 am
Issue 1:

5^k - 5^(k-1) = 5^k - (1/5)5^k

Is this true because regardless of the value of k, a negative one in the power means that it is at least divided by 5 once?

So would 5^(k+1) = 25^k ?

Issue 2:

ax^2 + bx = c = 0

When we use the quadratic formula to find x, we end up with two roots, right? Are there any instances where you just know beforehand that there is only one root?
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by DanaJ » Sat May 02, 2009 11:28 pm
1. Indeed, 5^k - 5^(k-1) = 5^k - (1/5)*(5^k). What you're actually doing here is spotting a common factor, which is 5^k. You can easily factorize it now, since it will be obvious that:

5^k - (1/5)*(5^k) = (5^k)*(1 - 1/5) = (5^k)*4/5

BUT

you made a mistake here:

5^(k+1) = 25^k

5^(k+1) = 5*(5^k) - as you can see, this is very different from 25^k. Just consider k = 3: 5^(k+1) = 5^4 = 625 = 5*125 = 5*5^3. HOWEVER, 25^4 is actually 625*625, since 25^2 = 625.


2. It's actually the easiest thing to know that the equation has only one root. This only happens when you can make a perfect square out of it, in other words when it follows the pattern on the formula:

(x + n)^2 = x^2 + 2xn + n^2.

Let me give you a few examples:
x^2 + 4x + 4 = x^2 + 2*2*x + 2^2 = (x + 2)^2 - the only root will be -2

x^2 + 10x + 25 = (x + 5)^2 - the only root will be -5

x^2 - 6x + 9 = (x - 3)^2 - the only root will be 3

x^2 - 4x + 3 - we can't find a square here, because x^2 - 4x + 3 = (x - 1)(x - 3) - as you can see, the equation has two roots, 1 and 3.
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