Question

Find the consecutive even positive integers whose squares have the sum 340.

Answer 1

SOLUTION :

Given : Sum of the squares of positive even integers  = 340

Let x and (x + 2) be the  two positive consecutive even integers.

A.T Q

x² + (x + 2)² = 340

x² + x² + 4x + 4 - 340 = 0

[(a + b)² = a² + b² + 2ab ]

2x² + 4x - 336 = 0

2(x² + 2x - 168) = 0

x² + 2x - 168 = 0

x²  + 14x - 12x - 168 = 0

[By middle term splitting method]

x(x + 14) - 12(x + 14) = 0

(x + 14)(x - 12) = 0

(x + 14) = 0 or (x - 12) = 0

x = - 14 or x = 12

Since, x is a positive number, so x ≠ - 14. Therefore x = 12

First integer = x = 12

Second integer = (x + 2) = 12 + 2 = 14  

Hence, the two positive consecutive even integers are 12 & 14 .

HOPE THIS  ANSWER WILL HELP YOU…..

Answer 2

Set x equal to the lower number, and x+2 as the higher number since they are consecutive even numbers so they are two apart.

Now write out the equation according to the question

(x)^2+(x+2)^2=340

x^2+x^2+4x+4=340

Combine like terms.

2x^2+4x+4=340

Set equal to zero so you can factor.

2x^2+4x−336=0

(2x+28)(x−12)=0

x=−14,12

x=12 because the answer must be positive according to the question.

That means x+2 is 14.

You can double check:

(12)^2+(14)^2=340

144+196=340

340=340