Original Question

You may be surprised to learn that the equation $(x^2-5x+5)^{x^2+4x-60}=1$ has five solutions.

Determine all five values of $x$ that satisfy the equation.

Attempts

At first I was trying to solve this as a polynomial, moving “1” from the RHS to the LHS, but then I did not know what to do. This is way more complicated to solve and I did not even know where to start.

Failing to solve this on my own, I put the whole function into Desmos, and this is what it gives me:

Ok here I can see four roots, but where is the fifth? Is it a real number even? And also what is this?

Eventually I asked my tutor about this and when he pointed me the way out I realised that I was being really dumb.

So in this functions there are three cases that will results in the RHS being 1.

Case #1

index ​$=x^2+4x-60=0$
$(x+10)(x-6)=0$
$​x=6$ or $x=-10$

Case #2

base = $x^2-5x+5=1$
$x^2-5x+4=0$
$(x−4)(x−1)=0$
$x=1$ or $x=4$

Case #3

This is the root that is not showing in Desmos.

base = -1, index = even
So $x^2+4x-60$ must be a even number
base = -1, $x^2-5x+5=-1$
$x^2-5x+6=0$
$(x−2)(x−3)=0$
$x=2$ or $x=3$
when $x=2$, index = $2^2+4*2-60=4+8-60=-48$
when $x=3$, index = $3^2+4*3-60=9+12-60=-39$ ≠ even
$x=2$

Answer

So the final 5 roots?
$x=-10, 1, 2, 4, 6$

Hope you find this interesting!

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There is one question that asks to find the area of a triangle, given two sides and the angle in between. When using the formula of Area = 0.5 · a · b · sin(C), I wrote the formula then calculated the answer directly, skipping the step of substituting the values of a, b and C in. And I got … 0 out of the 2 marks. So the step of substituting values into a formula should always be written.

It is also no ok, surprisingly (to me), to skip steps when solving equations. When I got to the step 16k2 - 112k + 192 < 0, the next step I wrote was 3 < k < 4, skipping the step of factorising the quadratic expression, I got a mark off. It is necessary to simplify the left side of < into k2 - 7k + 12 then factorise it: (k + 3)(k - 4).

After these marks lost due to skipping workings, here are my silly mistakes due to misreading the questions or not reading the key instructions in the question.

No.1

The graph below shows the relationship between log₁₀x and log₁₀y. Find y in terms of x.

So I went finding the formula assuming that the graph was plotted with y against x … and lost half of the marks.

No.2

The function f is defined by f(x) = 63/(4x-1) for 1 ≤ x ≤ 16.
The graph y = f(x) is shown below.
Find the range of f.

So here through the graph I found that the graph seems to have a horizontal asymptote of y = 0 and therefore I wrote – range: {f|f∈ℝ, f>0}… And obviously I got 0 mark for writing this. I didn’t use anything given in the question other than the graph. The right way to find the range should be substituting the highest and lowest value of the domain into the function to find out the highest and lowest values of the range.

So the above mistakes takes 9 marks away and I was barely over the mark required for a “7”. It is too bad that I don’t get a second chance for doing a paper for the report in October, but hopefully when it comes to the mocks in November, I would be able to avoid these mistakes.

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