Question

find three positive consecutive integers such that the product of the smallest and the largest is 26 more than six times the middle integer

Answer 1

Answer:

8, 9, 10

Step-by-step explanation:

three positive consecutive integers

x, x + 1, x + 2

the product of the smallest and the largest is 26 more than six times the middle integer

x(x + 2) = 6(x + 1) + 26

Expand

x^2 + 2x = 6x + 6 + 26

Combine like terms

x^2 + 2x = 6x + 32

Subtract 6x + 32 from both sides

x^2 - 4x - 32 = 0

Factor

(Alternately use the quadratic formula or complete the sqaure)

(x - 8)(x + 4)

x = {-4, 8}

-4 is extraneous because the integers must be positive

x = 8

x + 1 = 9

x + 2 = 10

8, 9, 10

Answer 2

Answer:

$8,9,10$

Step-by-step explanation:

We can make this an equation and use the variable n to represent our numbers

$n-1(smallest)$

$n(middle)$

$n+1(largest)$

$(n-1)\cdot (n+1)=6n+26$

$n^2-1 = 6n+26$

$n^2 - 6n - 27 = 0$

After we factorize we get

$n=9$

since our smallest is $n-1$ so our smallest number is $8$ and largest is $10$

brainliest please :)