math

classic Classic list List threaded Threaded
2 messages Options
CTHook CTHook
Reply | Threaded
Open this post in threaded view
|

math

Write three consecutive positive integers such that twice the product of the first and third is 2 less than 12 times the second.
Proof Proof
Reply | Threaded
Open this post in threaded view
|

Re: math

Maggie wrote
Write three consecutive positive integers such that twice the product of the first and third is 2 less than 12 times the second.
x = 1st
x + 1 = 2nd
x + 2 = 3rd

2[x(x + 2)] = 12(x + 1) - 2   ←twice the product of 1st and 3rd is 2 less than 12 times 2nd
2(x² + 2x) = 12x + 12 - 2   ← distributive property
2x² + 4x = 12x + 10   ← distributive property
2x² - 8x - 10 = 0   ← subtracted 12x and 10
2(x² - 4x - 5) = 0   ← factored 2 out
2(x - 5)(x + 1) = 0   ← factored into two binomials
(x - 5)(x + 1) = 0   ← divided by 2
x - 5 = 0   or   x + 1 = 0   ← set each factor equal to zero
x = 5   or   x = -1   ← added 5 and subtracted 1
x + 1 = 6  
x + 2 = 7  

5,6,7 are the 3 consecutive positive integers

- Proof