# find three consecutive odd integers such that 3 times the product of the first and third exceeds the product of the first and second by 28. ?

### 10 Answers

- PhilipLv 63 weeks ago
Let 3 consecutive odd integers be 2n-3, 2n-1, 2n+1...(1).

Then 3(2n-3)(2n+1) = (2n-3)(2n-1) + 28, ie., (2n-3)(6n+3-2n+1) = 28, ie.(2n-3)(n+1) = 7, ie., 2n^2 -n -10 = 0, ie.,

(2n-5)(n+2) = 0. Clearly, (1) holds only when n = -2. Then

the 3 consecutive odd integers are -7, -5, -3.

Check: 3(2n-3)(2n+1) = 3(-7)(-3) = 63

(2n-3)(2n-1) + 28 = (-7)(-5) +28 = 35+28 = 63.

Source(s): My source is myself. - la consoleLv 73 weeks ago
a = 2n - 1 ← an odd number cannot be divided by 2

b = a + 2

c = b + 2

3 times the product of the first and third exceeds the product of the first and second by 28.

3ac = ab + 28

3.(2n - 1).(b + 2) = (2n - 1).(a + 2) + 28

3.(2n - 1).(a + 2 + 2) = (2n - 1).(2n - 1 + 2) + 28

3.(2n - 1).(2n - 1 + 2 + 2) = (2n - 1).(2n - 1 + 2) + 28

3.(2n - 1).(2n + 3) = (2n - 1).(2n + 1) + 28

3.(4n² + 6n - 2n - 3) = (4n² + 2n - 2n - 1) + 28

3.(4n² + 4n - 3) = (4n² - 1) + 28

12n² + 12n - 9 = 4n² - 1 + 28

8n² + 12n = 36

n² + (12/8).n = 36/8

n² + (3/2).n = 9/2

n² + (3/2).n + (3/4)² = (9/2) + (3/4)²

n² + (3/2).n + (3/4)² = (9/2) + (9/16)

[n + (3/4)]² = 81/16

n + (3/4) = ± 9/4

n = - (3/4) ± (9/4)

n = (- 3 ± 9)/4

First case: n = (- 3 + 9)/4 = 6/4 = 3/2 ← no possible

Second case: n = (- 3 - 9)/4 = - 12/4 = - 3 ← ok

a = - 7

b = - 5

c = - 3

- ComoLv 73 weeks ago
:-

Let integers be 2x + 1 , 2x + 3 , 2x + 5

3 ( 2x + 1) ( 2x + 5 ) = ( 2x + 1 ) (2x + 3) + 28

3 ( 4x² + 12x + 5 ) = 4x² + 8x + 31

8x² + 28x - 16 = 0

2x² + 7x - 4 = 0

( 2x - 1 ) ( x + 4 ) = 0

x = - 4 is an integer value

The three integers are ( -7) , (-5) , (-3)

- KrishnamurthyLv 73 weeks ago
Three consecutive odd integers are such that

3 times the product of the first and third

exceeds the product of the first and second by 28.

Let (2n – 1), (2n + 1), (2n + 3) be 3 consecutive odd integers.

We have 3(2n – 1)(2n + 3) – (2n – 1)(2n + 1) = 28

12n^2 + 12n – 9 – (4n^2 – 1) – 28 = 0

8n^2 + 12n – 36 = 0

2n^2 + 3n – 9 = 0

(2n – 3)(n + 3) = 0

n = –3 or n = 3/2

The three consecutive odd integers are -7, -5, -3.

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- JOHNLv 73 weeks ago
Let 2n -1, 2n + 1, 2n + 3 be 3 consecutive odd integers.

We have 3(2n -1)((2n + 3) – (2n - 1)(2n + 1) – 28 = 0

12n^2 + 12n – 9 – (4n^2 - 1) – 28 = 0

8n^2 + 12n – 36 = 0

2n^2 + 3n – 9 = 0

(2n - 3)(n + 3) = 0

n = -3 (the other solution, n = 3/2, is a non-integer

whinch in any case yields even solutions)

The numbers are -7, -5, -3.

- no sea naboLv 63 weeks ago
(2n-1)(2n+3)=3x

(2n-1)(2n+1)=x+28

n=-3 x=7

(-7)(-3)=3(7)

(-7)(-5)=(7)+28

LQQD

(-3),(-5),(-7)

- llafferLv 73 weeks ago
Three consecutive odd integers. I'll use:

(x - 2), x, and (x + 2)

So as long as "x" is odd, the other two will be as well.

Three times the product of the first and third:

3(x - 2)(x + 2)

Exceeds the product of the first and second by 28:

3(x - 2)(x + 2) = x(x - 2) + 28

We can simplify both halves and solve as a quadratic:

3(x² - 4) = x² - 2x + 28

3x² - 12 = x² - 2x + 28

2x² + 2x - 40 = 0

x² + x - 20 = 0

(x + 5)(x - 4) = 0

x = 4 and -5

Since we only want odd values for x we can throw out the 4 to get our answer to be:

-7, -5, and -3