If you have the sume of squares of the digits of a positive two-digit number is 20, the tens digit is 2 more than the units digit.
answer is 2^2 and 4^2 (?)
Yes, that looks right.
a b
a^2+b^2 = 20
a = b+2
(b+2)^2 + b^2 = 20
b^2 + 4 b + 4 + b^2 = 20
2 b^2 + 4 b -16 = 0
b^2 + 2 b - 8 = 0
(b-2)(b+4 = 0
b = 2
a = 4
so number is indeed
42
To solve this problem, we can start by representing the two-digit number as 10a + b, where 'a' is the tens digit and 'b' is the units digit.
According to the problem statement, the sum of the squares of the digits is 20. So, we can write the equation as:
a^2 + b^2 = 20
The problem also states that the tens digit is 2 more than the units digit. We can represent this statement as:
a = b + 2
Now, we can substitute the value of 'a' from the second equation into the first equation:
(b + 2)^2 + b^2 = 20
Expanding and simplifying the equation, we get:
b^2 + 4b + 4 + b^2 = 20
2b^2 + 4b + 4 = 20
2b^2 + 4b - 16 = 0
Divide the equation by 2 to simplify it further:
b^2 + 2b - 8 = 0
Now, we can factorize the equation:
(b + 4)(b - 2) = 0
From this, we get two possible values for 'b': b = -4 and b = 2. However, since we are dealing with a positive two-digit number, the units digit cannot be negative. Therefore, we discard the value b = -4.
Now, substitute the value of 'b' back into the equation a = b + 2 to find the value of 'a':
a = 2 + 2
a = 4
So, the tens digit 'a' is 4 and the units digit 'b' is 2.
Therefore, the original two-digit number is 42.
To check if our answer is correct, we can calculate the sum of the squares of the digits:
4^2 + 2^2 = 16 + 4 = 20.
As the sum is indeed 20, our answer is correct.