Question

We know the following about the numbers a, b and c:

(a + b)^2 = 9,
(b + c)^2 = 25, and
(a + c)^2 = 81.
If a + b + c ≥ 1, determine the number of possible values for a + b + c.

Answer 1

Explanation:

(a + b)² = 9

(b + c)² = 25

(a + c)² = 81

Taking the square root:

a + b = ±3

b + c = ±5

a + c = ±9

By adding these three equations together and dividing both sides by 2, we get the value of a + b + c.

Possible combinations for a + b + c such that the sum is greater than or equal to 1 are:

a + b + c = (-3 + 5 + 9)/2 = 11/2

a + b + c = (3 − 5 + 9)/2 = 7/2

a + b + c = (3 + 5 + 9)/2 = 17/2