Three consecutive odd integers are such that the square of the third integer is 1515 greater thangreater than the sum of the squ
ares of the first two. one solution is 11, 33, and 55. find three other consecutive odd integers that also satisfy the given conditions.
1 answer:
3 consecutive odd integers....x, x + 2, and x + 4
(x + 4)^2 = (x^2 + (x + 2)^2) + 15
x^2 + 8x + 16 = x^2 + x^2 + 4x + 4 + 15
x^2 + 8x + 16 = 2x^2 + 4x + 19
2x^2 + 4x + 19 - x^2 - 8x - 16 = 0
x^2 - 4x + 3 = 0
(x - 3)(x - 1) = 0
x - 3 = 0.........x + 2 = 3 + 2 = 5.....x + 4 = 3 + 4 = 7
x = 3
so another one that satisfies this is : 3,5,7 <===
x - 1 = 0 x + 2 = 1 + 2 = 3......x + 4 = 1 + 4 = 5
x = 1
and another one is : 1,3,5 <===
You might be interested in
Answer:
D
Step-by-step explanation:
You can infer to the solution of this problem by looking at the choices. Hope this helped :)
Answer:
y
=
−
2
x
3
+
1
3
y
=
−
3
x
5
+
1
5
Step-by-step explanation:
24/12 < number of hours < 36/12
2 < number of hours < 3
Cheers and good luck!
Answer:
B
Step-by-step explanation: