Submitted by Jeff Golland on Wed, 06/28/2017 - 12:04
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Warm-up question 1 (iv) (a): wouldn't it be better to expand and simplify (2+root2)squared (rather than (1 + root2)squared) since that would be more relevant to part (b)?
Submitted by Jeff Golland on Wed, 06/28/2017 - 12:04
Warm-up question 1 (iv) (a): wouldn't it be better to expand and simplify (2+root2)squared (rather than (1 + root2)squared) since that would be more relevant to part (b)?
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Part b
I don't think it is more relevant to part (b). When I solve for $x^2$ one of my solutions is of the form
$$ x^2 = k \times (1+\sqrt{2})^2$$.
Warm down
I found different possibilities for bishops age and the ministers accordingly. However i didnt understand how to eliminate the others to reach the correct age. I found that minister can be 26, 36 or 50.
Warm down
I found that the minister can be 50 or more but not too much more as he has to look of a similar age with the bishop who is 32.
Warm down
The important thing to remember is that the Bishop knows how old he/she is! So if the Bishop was 42 they would be looking for the sum of ages which sums to 84.
Start by writing down 2450 as a product of primes. You can they write down all the possible sets of ages for the three bellringers (just to be nice, I'll tell you that none of them are ages 1). You then write down the sum of their ages. The fact that the bishop cannot say what the answer is immediately indicates which are the possible sets of ages.
I will put some more explanation in the hints and partials solution document to this question!
Warm down
There is now a full solution to this problem in the "Hints" document.
Q3, part ii
STEP question 3ii, part one.
I heard you found the discriminant and put it equal to 0? I did that, but still didn't know how exactly to prove anything. I substituted in the value for c^2 and c. Since the value for c is imaginary, I simply took it out of the equation due to lack of any other component. After doing this, I kept rearranging and rearranging. I finally got to the equation: (a(b^3)) - 2(a^2)(b^2) - 4ab + a^3. I've made a cubic factorization of three brackets, each containing x + [arbitrary greek letter] and then I expanded that and compared like terms with the equation I had. The only problem is, it's still complicated as all hell. I've probably done two whole sides of pure algebra just to get this far and I'm worried I've made a tiny mistake along the way.
Any help?
c isn't imaginary?
You are told in the "stem" of the question that $c$ is real!
You should have found a quadratic equation in $x$ [by multiplying throughout by $(x-a)(x-b)$]. It gets a bit messy but you can cancel some bits. You then find the discriminant of this and set it equal to 0. Again this gets a bit messy but you should be able to use this to show the required result $c^2=-\dfrac{4ab}{(a-b)^2}$. Note that since $c$ is real then we must have exactly one of $a$ and $b$ negative ($a,b$ are non-zero - given in the stem).
For the next "show that" it is probably easiest to start from the second condition and "work backwards". In an exam I would then either re-write it the correct way around or put something like "each of these steps can be reversed" or something similar!
The final part uses $c^2=1-\left( \dfrac{a+b}{a-b} \right)^2$ to show that $0 \lt c^2 \le 1$. You will need to use the fact that $a$ and $b$ are distinct and non-zero.