Question

Find three positive consecutive integers such that the sum of the fourth power of the smallest integer and eight times the middle integer is equal to two times the square of the largest integer increased by 63.

Answer 1

The numbers are 3 , 4 , 5

Explanation:

- Consecutive numbers are the number after each other like 1 , 2 , 3, .....

- We have three consecutive numbers

- Assume that the smallest is x, the middle is x + 1 and the largest

is x + 2

- The fourth power means power 4

- The fourth power of the smallest means
`x^(4)`

- Eight times the middle number means 8(x + 1)

- The sum of the fourth power of the smallest integer and eight

times the middle integer means
`x^(4)+8(x+1)`

∴ The sum of the fourth power of the smallest integer and eight

times the middle integer =
`x^(4)+8(x+1)` ⇒ (1)

- Two times the square of the largest integer means 2(x + 2)²

- Increased by 63 ⇒ means add 63

∴ Two times the square of the largest integer increased by 63 =

2(x + 2)² + 63 ⇒ (2)

- The sum of the fourth power of the smallest integer and eight times

the middle integer is equal to two times the square of the

largest integer increased by 63

∵ is equal means =

- Equate (1) and (2)

∴
`x^(4)+8(x+1)` = 2(x + 2)² + 63

- Lets simplify it

∵
`x^(4)+8x+8=2(x^(2)+4x+4)+63`

∴
`x^(4)+8x+8=2x^(2)+8x+8+63`

- Subtract 8 and 8x from both sides

∴
`x^(4)=2x^(2)+63`

- Subtract 2x² and 63 from both sides

∴
`x^(4)-2x^(2)-63=0`

- Now use your calculator to find the value of x

- The values of x are -3 , 3 , i√7 , -i√7

∵ The numbers are positive integers

∴ The value of x is 3

∴ The numbers are 3 , 4 , 5