half of the product of two consecutive numbers is 105. which equation can be used to solve for n, the…

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$n^{2}+n - 210 = 0$\n$n^{2}+n - 105 = 0$\n$2n^{2}+2n + 210 = 0$\n$2n^{2}+2n + 105 = 0$

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$n^{2}+n - 210 = 0$\n$n^{2}+n - 105 = 0$\n$2n^{2}+2n + 210 = 0$\n$2n^{2}+2n + 105 = 0$

Answer

Explanation:

Step1: Define the consecutive numbers

Let the smaller number be $n$, then the larger consecutive number is $n + 1$.

Step2: Write the product expression

The product of the two consecutive numbers is $n(n + 1)$.

Step3: Set up the equation based on the problem

Half of the product is 105, so $\frac{1}{2}n(n + 1)=105$.

Step4: Simplify the equation

Multiply both sides by 2 to get $n(n + 1)=210$. Expand the left - hand side: $n^{2}+n = 210$. Rearrange to the standard quadratic form: $n^{2}+n - 210=0$.

Answer:

$n^{2}+n - 210 = 0$