I need help with an algebra question involving consecutive odd integers and multiplication.?

Question

Find three consecutive integers for which the product of the first and the second is 377 less than 40 times the third.

And could you please show how you solved it. It will help me in the future : )

Answer 1

Ooh, I love these. OK,

a* (a+1)= 40(a+2)-377

that's the word problem restated.

So a^2 + a = 40a +80 -377

or

a^2 -39a +297 = 0

Now use the quadratic equation to solve

___________________

I can't tell if you meant consecutive, or "consecutive odd". If it's consecutive odd, then (a+1) and (a+2) get replaced by (a+2) and (a+4) and you continue on from there.

Answer 2

OK. Since your question indicates these are consecutive ODD integers (and since no 3 consecutive integers work) I'll assume you meant odd integers. The 3 odd integers are x, x+2, and x+4 since you must add 2 to an odd integer to get another odd integer.

So you have x(x+2) = 40(x+4) - 377

This is x^2 + 2x = 40x + 160 - 377

Moving all to the left you get x^2 + 2x - 40x - 160 + 377 = 0
which simplifies to x^2 - 38x + 217 = 0

By FOIL this is (x-7)(x-31) = 0

So x could be 7 or it could be 31, since both are odd. Then go from there.

Answer 3

Let the consecutive odd integers be x-2,x and x+2
Therefore according to the problem,
x(x-2)=40(x+2)-377
=>x^2-2x=40x+80-377
=>x^2-2x-40x-80+377=0
=>x^2-42x+297=0
=>x^2-9x-33x+297=0
=>x(x-9)-33(x-9)=0
=>(x-9)(x-33)=0
Therefore x=9 or 33
Therefore the numbers are 7,9,11 or 31,33,35.

Answer 4

Let the numbers be n, n+1, n+2.

n(n+1) +377 = 40*(n+2)

n^2 + n + 377 = 40n + 80

n^2 -39n + 297 = 0. But this doesn't have integer solutions.

Answer 5

Surely you mean three consecutive odd integers?
They are:
x, x + 2, and x + 4
The equation becomes:
x(x + 2) = 40(x + 4) - 377
Simplify and rearrange We get:
x^2 - 38x + 21217 = 0
Factoring we get:
(x - 7)(x - 31) = 0
x = 7; x = 31
The two solutions are 7, 9, 11; and 31, 33, 35

Answer 6

ok, i do not recognize each and each of the steps and stuff. yet it truly is how I figured it out. First, I divided fifty six with the help of four. Then, I have 14 because the overall. So, I positioned it as a mean and grew to grow to be it into magnificent numbers. One a lot less of 14 is 13. an extra advantageous of 14 is 15. 3 a lot less of 14 is 11 and three more advantageous of 14 is 17. So the solutions is 11, 13, 15, 17. desire it help. Sorry, i do not bear in mind a thanks to do it.

Answer 7

Geeezzzzz Fcas80, that's because it was consecutive -odd- integers ☺

Doug