Question

Find 4 consecutive odd integers where the product of the two smaller numbers is 64 less than the product of the two larger numbers. @johnweldon1993

Answer 1

Can you apply what I did in your last question to this one?

Answer 2

idk but it wont be the same answer thought

Answer 3

cuz they want 4 consecutive for this problem

Answer 4

that's correct....this time it is odd numbers....we can think of this as, whatever even number you have...when you add an odd number to it...you'll have an odd number right?

So here you'll have

2n + 1, 2n + 3, 2n + 5, and 2n + 7

this will give you 4 numbers...now can you use those old steps to solve it?

Answer 5

Here I'll do the equation again for you and see if you can continue it

product of the 2 smaller numbers is 64 less than the product of the 2 bigger numbers

(2n + 1)(2n + 3) = (2n + 5)(2n + 7)

Answer 6

Woops....use this new equation

(2n + 1)(2n + 3) = (2n + 5)(2n + 7) - 64

Answer 7

3,5,7,9 i this the correct answer

Answer 8

We can check your answer

the product of the 2 smaller numbers you have (15) is 64 less than the product of the 2 larger numbers (63) ...so no that cannot be correct...try out my equation

Answer 9

(2n + 1)(2n + 3) = (2n + 5)(2n + 7) - 64

\[4n^2 + 8n + 3 = 4n^2 + 24n + 35 - 64\]

can you continue on?

Answer 10

can u gib=ve me the answer plz

Answer 11

I cannot give you the answer...I can show you how to do it again for future reference

Answer 12

\[8n + 3 = 24n - 29\]
\[-16n = -32\]
\[n = 2\]

now that you know 'n' = 2

you can go back to those original numbers we had

2n + 1, 2n + 3, 2n + 5, 2n + 7 and replace all the 'n' with 2

2(2) + 1 = 5
2(2) + 3 = 7
2(2) + 5 = 9
2(2) + 7 = 11

Answer 13

now we can check

product of 2 smaller (35) is 64 less than product of 2 larger (99)

99 - 35 = 64 so those numbers are correct