For the first equation:
1.According to the rules of integers, when the exponent is 0, the answer will always be 1.Therefore, for the 1st equation, we can take (n^2 -11n+30)=0. This is an example of the quadratic equation : (n-a)(n-b). Therefore we will have to find the values of a and b. When you expand and substitute, you will find that a = 5 and b = 6. Therefore, n will be equal to either 5 or 6. These are the first 2 solutions for the equation.
2. Again, according to the rules of integers, when the base of any equation is 1, the solution will also be 1 irrespective of any power to that base. Therefore, we can take (n^2-5n+5) = 1. To make this a quadratic equation, we have to make the right hand side of the equation = 0. To do this, we have to bring the 1 to the left hand side. The equation will now be (n^2-5n+5-1)=0.
n^2-5n+4=0
(n-a)(n-b)
Therefore, a = 4 and b = 1
(n-4)(n-1)=0
Therefore n is equal to either 4 or 1
These are the 3rd and 4th solution to the equation.
3. When the base of any equation is -1 and the exponent is equal to an even number, the answer will always 1. Therefore the base of the equation can be equated to -1 :
n^2-5n+5 = -1
n^2-5n+5+1 = 0
n^2-5n+6 = 0
(n-a)(n-b)
a = 3 and 2
(n-3)(n-2) = 0
Therefor n will be equal to either 3 or 2.
However, for these solutions to be correct, we have to check if the exponent will be even.
n^2-11n+30
if n = 3, then
9-33+30
=6
If n = 2 then
4-22+30
=12
Therefore these will be our final 2 solutions.
The 6 solutions for the first equation are n = 5,6,1,4,3,2
For the second equation, using same steps as above :
1. n^2-13n+42 = 0
(n-a)(n-b)
a=7,b=6
(n-7)(n-6) = 0
Therefore n can either be 7 or 6
These are the first 2 solutions for the equation
2. n^2-7n+11 = 1
n^2-7n+11-1 = 0
n^2-7n+10 = 0
(n-a)(n-b)
a = 5, b = 2
(n-5)(n-2) = 0
Therefore n can either be 5 or 2
These are the 3rd and 4th solutions to our equation
3. When the base of any equation is -1 and the exponent is equal to an even number, the answer will always 1. Therefore the base of the equation can be equated to -1 :. Therefore:
n^2-7n+11 = -1
n^2-7n+11+1 = 0
n^2-7n+12 = 0
(n-a)(n-b)
a = 3 and 4
(n-3)(n-4) = 0
There n will equal to either 3 or 4
Now we have to check if the exponent will be even or not.
n^2-13n+42
if n = 3, then
9-39+30
=12
If n = 4 then
16-42+42
=18
Therefore these will be our final 2 solutions.
Thus the 6 solutions to the second equation is n = 7,6,5,2,3,4