the product of two consecutive positive integers is 120?

Answer 1

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The product of two consecutive positive integers is 120. Find the positive integers.

Solution:
n • (n + 1) = 120
(n • n) + (n • 1) = 120
n² + n = 120
n² + n - 120 = 0

Use the quadratic formula:
-b ± √( b ² - 4ac)
-----------------------
2a

a = 1, b = 1, c = -120

-1 ± √( 1 ² - 4(1)(-120))
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2(1)

-1± √(481)
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2

n1 =~10.45 and n2 = ~ -11.45

Since one of these answers is negative, and neither is an integer, there are no solutions, no two consecutive integers whose product is 120. <== ***Answer***
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Answer 2

The PRODUCT of consecutive integers n and n+1 is n^2 + n. So the problem is equivalent to solving n^2 + n = 120. But you can check that n^2 + n - 120 = 0 has no integer solutions (10^2 + 10 - 120 is negative, 11^2 + 11 - 120 is positive), which means that no two consecutive integers have product 120.

Answer 3

ok...if our first number is x, then the next number would be x+1. Product means multiplication so
x(x+1) = 120
x^2 + x -120 = 0

hmmm...did you mean the product of two consecutive EVEN positive integers is 120?

no 2 consecutive integers = 120 when multiplied. remember an integer is a whole number...

i am confused...sorry...can't help

Answer 4

I believe that the fact that you end up with x^2 + 2X - 120 = 0, whose solutions are -12 and 10, means that there are no consecutive even numbers whose product is 120. 10 and 12 are even and consecutive, but 10 and -12 are even but not consecutive, which proves there are no two consecutive even numbers satisfying those conditions.

Answer 5

Make the first positive integer x. In order to get the next consecutive positive, add 2. (Think of it as x being 2...You need to add 2 more to get 4, the next positive integer.)

You take the first integer, x, and multiply it by the next positive integer, x+2.

x(x+2)=120
x^2 + 2x=120

From here, you would either use the quadratic formula, complete the square, or just factor.

Completing the square would give you...

(x+1)^2 = 121
x+1=11
x=10

If x equals 10, the first integer is 10, and the second integer is 10+2, which is 12.

Final answer: 10 and 12

Answer 6

x(x + 1) = 120
x² + x = 120
x² + x - 120 = 0

Using the quadratic formula

x = [-1 ± √(1² - 4*1*-120)]/2
x = (-1 ± √481)/2

Since √481 is not an integer, x is not an integer, so two such integers do not exist.

Answer 7

n^2 + n -120 = 0 and use the quadratic equation.

Skylor Williams

Answer 8

I am not sure what you are asking. There are no two intesgers who's product is 120. If there were, what are you asking about them?

Answer 9

it's not x+x+1 like in the last answer, it's the product, not the sum
x(x+1) = 120
x^2 + x = 120
x^2 + x - 120
use quadratic formula

Answer 10

n(n+1) = 120

n^2-n-120 = 0

solve for n