Half of the product of two consecutive numbers is 105. Which equation can be used to solve for n, the smaller of the two numbers?

n^2 + n - 210 = 0

n^2 + n - 105 = 0

2n^2 + 2n + 210 = 0

2n^2 + 2n + 105 = 0

Question

Mathematics

Guest

14 December, 20:53

Half of the product of two consecutive numbers is 105. Which equation can be used to solve for n, the smaller of the two numbers?

n^2 + n - 210 = 0

n^2 + n - 105 = 0

2n^2 + 2n + 210 = 0

2n^2 + 2n + 105 = 0

+1

Answers (1)

Know the Answer?

Cello · Answer 1

Hello:

the equation is:

n^2 + n - 105 = 0

because you have : n²+n = n (n+1) ... (product of two consecutive numbers)

so: n (n+1) - 105 = 0

conclusion : n (n+1) = 105

Half of the product of two consecutive numbers is 105. Which equation can be used to solve for n, the smaller of the two numbers?n^2 + n - 210 = 0n^2 + n - 105 = 02n^2 + 2n + 210 = 02n^2 + 2n + 105 = 0

Half of the product of two consecutive numbers is 105. Which equation can be used to solve for n, the smaller of the two numbers?

n^2 + n - 210 = 0

n^2 + n - 105 = 0

2n^2 + 2n + 210 = 0

2n^2 + 2n + 105 = 0