are their number with property that the sum of their square of their sum
Answers
A non-trivial solution is (a,b,c)=(1, 2, -2/3).
(a+b+c)^2 = (7/3)^2 = 49/9
a^2 + b^2 + c^2 = 1+4+4/9 = 49/9.
Or restricted to integers, (a,b,c)=(2, 2, -1); squares sum to 9.
I’ll guess that distinct non-zero integers is also possible but requires more than 3 numbers.
PS - Characterizing what is possible with three numbers is easy.
Expanding the square of a sum and deducting the sum of a square leaves a sum of mixed terms that must sum to 0 to satisfy the problem. In the case of 3 numbers a, b, c:
0 = 2(ab+ac+bc) = 2[ab+c(a+b)]. Hence the problem is equivalent to solving
-ab = c(a+b). Allowing non-integers, that’s trivial. Just choose any non-zero values for a,b, then put c=-ab/(a+b), which is what I did before. Clearly any choices for a,b are fine & compute c from the formula but c may not be an integer. As for integer solutions it is sufficient that a & b are integers such that a+b=-1 — i.e. sequential numbers except of opposite sign. To have a solution in distinct integers, insist further than a & b are at least 2 in absolute value. Example: a=-2, b=3, c=6.
Sum is 7 which squares to 49, and 49=4+9+36.
So contrary to my guess of no solutions in three distinct integers, there are infinitely many. Well, I was close.
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Step-by-step explanation:
let a and b be the two numbers such that the sum of their squares is the square of their sum.
=>
simplifying we get,
2ab = 0
=> ab = 0
either a = 0 or b = 0
thus the two numbers are 0 and b where b is any real number such that their sum of squares is the square of their sums