What is three consecutive numbers whose square add to 245

Question

Find the three consecutive numbers whose square add to 245

Answer 1

Three consecutive numbers can be written as:

x, x+1, x+2

 

We know their squares add up to 245.  Therefore,

 

x² + (x+1)² + (x+2)² = 245

 

Expanding the terms:

x² + x² + 2x + 1 + x² + 4x + 4 = 245

 

Combine like terms:

(1 + 1 + 1)x² + (2 + 4)x + 1 + 4 = 245

3x² + 6x + 5 = 245

3x² + 6x -240 = 0

Divide through by 3:

x² + 2x - 80 = 0

(x + 10)(x - 8) = 0

X = -10 or x =8

 

You'll have 2 sets of answers.

(-10, -9, -8)  and (8, 9, 10)

 

 

Answer 2

Here's a little trick using arithmetic.

245/3=81.666...

√81.666...=9.036

8,9,10  are the answers as well as -10,-9,-8

another example: 3 consecutive integers whose squares sum to 509

509/3=169.666...

√169.666=13.025

12,13,14 are the answers

144+169+196=509

Answer 3

Hello, Lisa

Let x       be the first number

Let x+1   be the second number

Let x+2   be the third consecutive number

 

So,

x^2 + (x+1)^2 +(x+2)^2= 245

x^2 +[x^2+2x+1]+ [x^2+4x+4]= 245         Combining like terms

           3X^2+ 6x+5                    = 245        Subtracting 5 from both sides

                           -5                        -5   

             3X^2+6x                       =240          Divide all terms by 3

                                 x^2+2x     = 80           Subtracting 80 from both sides

                                           -80    -80

                      x^2+2x-80= 0                       Factoring

                     (x+10)(x-8)= 0                        

 x=-10 or x=8

 

So, the numbers could be 8,9,10  or -10,-9, -8 .  (64)+(81)+(100)= 245

 

I hope this helps out and thanks for posting.

 

D.Y. Taylor

Don't forget to rate me :)

           

Answer 4

Let x = the smallest of the three integers.  The others are x+1 and x+2.

 

x² + (x+1)² + (x+2)² = 245

x² + x² + 2x + 1 + x² + 4x + 4 = 245

3x² + 6x + 5 = 245

3x² + 6x -240 = 0

 

x² + 2x - 80 = 0

 

(x+10)(x-8)

 

x = -10 and 8

 

There are two solutions:  -10, -9, -8 and 8, 9, 10

Answer 5

This problem can also be solved using algebra.

Let X = the lowest of the three numbers so x + 1 is the second number and x + 2 is the third number.

Then you can write the equation:

x^2 + (x+1)^2 + (x+2)^2 = 245

X^2 + (x+1)(x+1) + (x+2)(x+2) = 245

This becomes:

x^2 + x^2 + 2x + 1 + x^2 + 4x + 4 = 245

This can be simplified by adding like terms and subtracting 245 from both sides to get:

3x^2 + 6x - 240 = 0

3(x^2 + 2x - 80) = 0

The trinomial can be factored to get:

3(x+10)(x-8) = 0

Therefore x must be either -10 or 8.

 

Assuming you want positive numbers the correct answers would be 8, 9, and 10, but -10, -9, and -8 would also work.

Answer 6

This is a trial and error.

 

try 8, 9, 10

 

8² + 9² + 10²

 

64 + 81 +100 = 245

 

so, consecutive numbers are 8, 9, 10.