Question

the sum of the squares of 3 consecutive odd numbers is 83 find the number

Answer 1

let the first no be x so the other two will be x+1,x+2

x^2+(x+1)^2+(x+2)^2=83

quadratic equation would be formed do it by iddle term splitting....

sorry i cant do it now

Answer 2

Answer:

Step-by-step explanation:

Given the sum of the squares of 3 consecutive odd numbers is 83

We have to find out the numbers

Let the first number be 'x' , Second number be '(x+2)²'

And third number be '(x+4)²'

$\boxed{\bold{x^2 + (x+2)^2 + (x+4)^2 = 83}}$

We know that,

(x+2)² = x² + 4x + 4

(x+4)² = x² + 8x + 16

x² + x² + 4x + 4 + x² + 8x + 16 = 83

3x² + 12x + 20 = 83

3x² + 12x - 63 = 0

3( x² + 4x - 21 ) = 0

x² + 4x - 21 = 0

Solve this quadratic equation

=  $\bold{\frac{-b\frac{+}{-}\sqrt{b^2\:-\:4ac} }{2a} }$

= $\bold{\frac{-4\frac{+}{-}\sqrt{16\: +\: 4\:*\:1\:*\:-\:21} }{2} }$

= $\bold{\frac{-4\frac{+}{-}\sqrt{100} }{2} }$

= $\bold{\frac{-4\frac{+}{-}10 }{2} }$

= $\bold{\frac{-4+10 }{2} }\: , \:\bold{\frac{-4-10 }{2} }$

= $\bold{\frac{6 }{2} }\: , \:\bold{\frac{-14 }{2} }$

⇒ x = 3

Hence the numbers are 3 , 5 , 7

3² + 5² + 7² = 9 + 25 + 49 = 83