product of three consecutive integers.....

[irene12]

Hello,I don't remember exactly how to finish
The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.
(x)(x+2)(x+4)-23=(x+2)3-99
(x)(x+4)-23=(x+2)2-99
2x+4x=(x+2)2-76
6x=x2+4-76
x=x2/6-72/6
x=x2/6-12
x=x2-2
I don't think that i got it quite right?

 

[Deleted member 4993]

Hello,I don't remember exactly how to finish
The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.
(x)(x+2)(x+4)-23=(x+2)[sup:3ujgk6d7]3[/sup:3ujgk6d7]-99
(x)(x+4)-23=(x+2)2-99 >>>>>>>>>>>>>> Incorrect step
2x+4x=(x+2)2-76
6x=x2+4-76
x=x2/6-72/6
x=x2/6-12
x=x2-2
I don't think that i got it quite right?

(x)(x+2)(x+4)-23=(x+2)[sup:3ujgk6d7]3[/sup:3ujgk6d7]-99

(x)(x+2)(x+4) - (x+2)[sup:3ujgk6d7]3[/sup:3ujgk6d7] = 23 - 99

(x+2)[(x+2)[sup:3ujgk6d7]2[/sup:3ujgk6d7] - x(x+4)] = 76

(x+2)[x[sup:3ujgk6d7]2[/sup:3ujgk6d7] + 4x + 4 - x[sup:3ujgk6d7]2[/sup:3ujgk6d7] - 4x] = 76

Now continue....

 

[JeffM]

Hello,I don't remember exactly how to finish
The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.
(x)(x+2)(x+4)-23=(x+2)3-99 Use the carat symbol (shift + 6) for exponentiation (x + 2)^3.
(x)(x+4)-23=(x+2)2-99
2x+4x=(x+2)2-76 How do you get (2x + 4x) = x(x + 4)?
6x=x2+4-76 How do you get (x + 2)[sup:umxme5ne]2[/sup:umxme5ne] = x[sup:umxme5ne]2[/sup:umxme5ne] + 4?
x=x2/6-72/6
x=x2/6-12
x=x2-2 How do you get (x[sup:umxme5ne]2[/sup:umxme5ne] / 6) + 12 = (x[sup:umxme5ne]2[/sup:umxme5ne] + 2)?
I don't think that i got it quite right? No. You understand how to do it, but you made some careless errors. You can do this easily if you take care

 

[Deleted member 4993]

Hello,I don't remember exactly how to finish
The product of three consecutive odd integers reduced by 23 is 99 less than the cube of the sum of the smallest number and 2. Compute the mean of the three integers.
(x)(x+2)(x+4)-23=(x+2)[sup:2jlksm1n]3[/sup:2jlksm1n]-99
(x)(x+4)-23=(x+2)2-99 >>>>>>>>>>>>>> Incorrect step
2x+4x=(x+2)2-76
6x=x2+4-76
x=x2/6-72/6
x=x2/6-12
x=x2-2
I don't think that i got it quite right?

(x)(x+2)(x+4)-23=(x+2)[sup:2jlksm1n]3[/sup:2jlksm1n]-99

(x)(x+2)(x+4) - (x+2)[sup:2jlksm1n]3[/sup:2jlksm1n] = 23 - 99

(x+2)[(x+2)[sup:2jlksm1n]2[/sup:2jlksm1n] - x(x+4)] = 76

(x+2)[x[sup:2jlksm1n]2[/sup:2jlksm1n] + 4x + 4 - x[sup:2jlksm1n]2[/sup:2jlksm1n] - 4x] = 76

Now continue....

(x+2)[x[sup:2jlksm1n]2[/sup:2jlksm1n] + 4x + 4 - x[sup:2jlksm1n]2[/sup:2jlksm1n] - 4x] = 76

(x+2)*4 = 76

x+2 = 19 ? x = 17

Hence numbers are 17, 19 & 21

The mean of these numbers is 19

check

17 * 19 * 21 - 23 = 6783 -23 = 6760

19[sup:2jlksm1n]3[/sup:2jlksm1n] - 99 = 6859 - 99 = 6760

Checks