Question

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?

n2 + n – 210 = 0
n2 + n – 105 = 0
2n2 + 2n + 210 = 0
2n2 + 2n + 105 = 0

Answer 1

Answer:

n² + n -210 = 0

LET THE FIRST NUMBER BE n

THEREFORE ITS CONSECUTIVE IS NUMBER BE n+1

ACCORDING TO THE CONDITION

½n.(n+1) = 105

n(n+1) = 210

n² + n = 210

n² + n - 210 = 0

Answer 2

The answer is $n^2 + n - 210 = 0$

Step-by-step explanation:

Assuming one number to be 'n' and then the other would be (n+1)

Then, their product would be $(n)(n+1)$

Half of the product would be  $\frac{1}{2}(n)(n+1)$

Thus, the equation would be $\frac{1}{2}(n)(n+1) = 105$

Cross multiplying, we obtain $(n)(n+1) = 2\times105$

Opening the brackets $n^2 + n = 210$

Bringing '210' on the left-hand side, we obtain $n^2 + n - 210 = 0$

Hence, the final answer is the first option.