Question

find two consecutive odd +ve integer sum of whose squares is 74​

Answer 1

the numbers be x and x+1

according to question

x2+(x+1)2=74

x2+x2+2x+1=74

2x^2+2x-74+1=0

2x^2+2x-73=0

Answer 2

"TWO CONSECUTIVE ODD POSITIVE INTEGERS!!!"

 

Isn't the difference between them 2?!

 

So let one be x and the other be x + 2.

 

WAIT!!!

 

If we take them as this,

   

we get a second degree equation,

   

then have to solve the problem by 'completing the square',

     

or by 'splitting the middle term',

   

or by "- b ± √b² - 4ac / 2a",......

     

OH NO! IT TAKES TOO MUCH TIME!!!

   

Who doesn't want to do it easier???!!!

   

Okay, let one be x - 1 and the other be x + 1.

   

NOW LET'S SEE!

   

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TO REMEMBER...

 

$\bold{(a - b)^2 + (a + b)^2 = 2(a^2 + b^2)}$

 

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So,

 

(x - 1)² + (x + 1)² = 74

   

2(x² + 1) = 74

   

x² + 1 = 37

   

x² = 36

   

x = ± 6

   

"'BUT  x ≠ -6!!! BUT WHY?! BUT WHY?!"'

   

∴ x = 6

   

x - 1 = 5

   

x + 1 = 7

 

""SO THE NUMBERS ARE '5' AND '7'!!!""

 

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Thank you. Have a nice day. :-)

   

     

   

#adithyasajeevan