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

A.) n^2 + n – 210 = 0
B.) n^2 + n – 105 = 0
C.) 2n^2 + 2n + 210 = 0
D.) 2n^2 + 2n + 105 = 0

Respuesta :

Answer:

A

Step-by-step explanation:

Consecutive nnumbers are spaced 1 units apart.

Such as 4,5,6,...

or 12,13,14,...

Thus,

if smaller number is n, then the next number (consecutive) is n+1

Since, half of the product of them is 105, we can write the equation as:

[tex]\frac{1}{2}((n)(n+1))=105[/tex]

We can do some algebra and make it in quadratic form as shown below:

[tex]\frac{1}{2}((n)(n+1))=105\\\frac{1}{2}(n^2 + n)=105\\n^2+n=2*105\\n^2+n=210\\n^2+n-210=0[/tex]

Answer choice A is right.

Answer:

A.) n^2 + n – 210 = 0

Step-by-step explanation:

just took edg 2022. hope this helps!

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