I tackled the first problem and I noticed that I could try and make the quadratic equation at the bottom to be equal to 1. Because, 1 times the power of any integer number is always going to be 1. So I made an equation like this (by the way n^2 is meant to be n squared):
n^2-5n+5=1
Next, I tried looking for possible values for n to be equal by 1, by taking 1 off each side:
n^2-5n+4=0
I realised you can factorise the equation on the left side, so I did it to get to the final step.
(n-1)(n-4)=0
From this equation, we can deduce that n=1 and n=4. So we have two solutions to the first problem. I was really excited! Next, I tried looking for ways for making the power quadratic equation to be equal to 0. Because a positive number with a power of 0 will always be equal to 1. So I made another algebraic equation.
n^2-11n+30=0
You can factorise this like the other equation to get this simplified equation:
(n-6)(n-5)=0
From this, n=6 and n=5. Now we have four solutions (Yay!), 1,4,5 and 6. I realised we had two more possible values left and I thought it couldn't be a negative number as it would result in a really high number with a high power. It couldn't be a fraction as it wouldn't be equal to a integer. I thought that maybe we could have something like -1 to the power of an positive even integer.Then it suddenly popped in my head that the two numbers I'm missing are 2 and 3. Because that would mean all the 6 values of n are in number order from 1 to 6. I might be onto something!
I tried substituting 2 as a value of n into the mega quadratic equation and I came up with -1 to the power of 12 which would come up with 1! I did the same thing with 3 and I got -1 to the power of 6! I had all 6 solutions for the first problem!
For the second question, I did the same thing and made equations for the bottom number to be equal to 1 and another one with the power equal to 0 and they appear respectively.
(n-2)(n-5)=0
(n-6)(n-7)=0
From these equations, n can be equal to 2,5,6 and 7. I noticed I was missing two numbers like last time and I could fill in a number sequence with 3 and 4. I got -1 to the power of 12 substituting 3 as n and -1 to the power of 6 substituting 4 which both equal to 1. Now I have found the answers for the second question!
In order to find a pattern for these mega quadratic equations I looked at how I found the values of n for both equations to find a pattern. I noticed that when I was making equations for the bottom number to be equal to 1, The values I found to be n in the first question was 1 and 4. The values in the second question was 2 and 5. The values of n were added once!
I did the same thing with the power quadratic equation and found in the first question and second question, I got 5,6 and 6,7 respectively. I tried finding relationships with the values of n with the mega quadratic equation and I found out that the values of n that I found by making the bottom number 1 helps make the numbers in the equation. Eg. The -5 and +5 in this equation.
(n^2−5n+5)
The values of n I found in the first question (1 and 4) make the numbers as they add up to make the 5 in -5n (the minus just stays there) and to get the 5 on the right you have to multiply the two values of n together and add 1. The reason you don't add them both together, is because in the second question I found out that 2+5 isn't going to make the 11 on the right side of the equation.
(n^2−7n+11)
But, 2 multiplied by 5 and adding one to the answer will get you the number on the right side of the quadratic equation on the bottom. This can be done for the first equation. I did something similar to the power quadratic equation and found that to get the number in the middle, you have to add the values of n you find while using the equation which makes the power quadratic equation equal to 0.
For the number on the right, you can multiply the two values of n together to get the number on the right side of the equation (you don't have to add 1). And for the number in the middle, you can add the numbers of n that you found when making the power equation equal to 0. From what I found, I made an equation to make a mega quadratic equation (I think I made an ultimate quadratic equation!).
(n^2 - (2x+3)n+(x(x+3)+1) to the power of (n^2 - (2x+9)n +(x+4)(x+5))
(x is an integer and the lowest possible value of n that will make the whole equation equal to 1)
Eg. The first question's values of n are 1,2,3,4,5 and 6. 1 is the lowest value of n that makes the equation equal to 1.
Using this equation, you can find an infinite amount of mega quadratic equations like:
(n^2−9n+19) to the power of (n^2−15n+36)=1
(n^2-11n+29) to the power of (n^2-17n+72)=1
You can find many mega quadratic equations and its answers by using the equation!