Submitted by nakapawa on Wed, 09/14/2016 - 18:47

I have done question 3 part i and i am happy with my answer. Not sure how to tackle part ii though. Any hints please? Thanks in advance.

Submitted by nakapawa on Wed, 09/14/2016 - 18:47

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I have done question 3 part i and i am happy with my answer. Not sure how to tackle part ii though. Any hints please? Thanks in advance.

## 3ii

Start in the same sort of way as you (probably) did part (i) - i.e. multipliy throughout by (x-a)(x-b) to remove the fractions. There are quite a few brackets to expand but you should end up with a quadratic in c. Then, as there is exactly one real solution, we have "$B^2 - 4AC=0$" (Note the use of capital letters). More expansion and simplification should lead to the required $c^2 = ...$.

Try this bit first, then ask if you need help on the next bit.

## 3ii

When you remove the fraction and end up with a quadratic in C, does that mean everything has a a C in it and if so would the substitution into "B^2-4AC" be enormous with loads of brackets to expand? That is what I did and I got a ridiculous sum of a mixture of As Bs and Cs.

## I also got to a point where i

I also got to a point where i subbed in for B^2 - 4AC and then wasn't really sure where to go from there. Not sure whether to sub in for C^2 as what it gives in the question or do something else. I got to a dead end where i got an answer C=1, which doesn't seems right.

## c/C

Be very careful not to confuse c and C!

With your quadratic equation in $x$ then coefficient of $x^2$ should be something like $(1-c)$. With "$B^2-4AC$" you should end up with something which has $c^2$ in it, but no $c$ term.

## deduce

I'm a little confused on the last bit, any advice? not really sure where to start

## Deduce

HINT: You have $c^2 = 1 - something^2$,

HINT: $c^2$ must be greater than or equal to zero (as $c$ is real, and something real squared cannot be negative), but we know that $c^2=-\dfrac{4ab} {(a-b)^2}$,

Ask again if you need more!

## I know that this means that a

I know that this means that a and b have different signs, unlike in the last part, but I couldn't see how that helps, since we get a different equation altogether- we don't have anything like x^2=bc.

## SPOILERS

Since $c^2=1-something^2$ we know that $c^2 \le 1$.

Since $c$ is real, we must have $c^2 \ge 0$. Also, we are told in "the stem" that $a$ and $b$ are "Non-zero, real numbers" which means that we have $4ab \neq 0$! Hence $c^2 > 0$ as it cannot be equal to 0.

## Last part of ii

How do you do the last part of question 3 ii?

## 3ii

Have you read the above comments on this thread? Try working through some of the hints above and then ask again if it still isn't coming out.