Question

Find two consequtive odd integers some of whose sq is 290

Answer 1

Let one of the odd positive integer be x 

then the other odd positive integer is x+2

their sum of squares = x^2 +(x+2)^2

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

                               = 2x^2 + 4x + 4

Given that their sum of squares = 290

⇒ 2x^2 +4x + 4 = 290

⇒ 2x^2 +4x = 290-4 = 286

⇒ 2x^2+ 4x -286 = 0 

⇒ 2(x^2 + 2x - 143) = 0 

⇒ x^2 + 2x - 143 = 0

⇒ x^2 + 13x - 11x -143 = 0 

⇒ x(x+13) - 11(x+13) = 0 

⇒ (x-11) = 0 , (x+13) = 0

Therfore , x = 11 or -13

We always take positive value of x 

So , x = 11 and (x+2) = 11 + 2 = 13 

Therefore , the odd positive integers are 11 and 13 .

Answer 2

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

Given:

• Sum of squares of 2 consecutive odd integers = 290

To Find:

• Value of the numbers

Assumption:

• Let the 2 consecutive odd integers be (2x+1) & (2x+3) respectively.

Solution:

According to the question,

$\displaystyle (2x+1)^2 + (2x+3)^2 = 290 \\ \\ We\:know\:that,\\ \\ (a+b)^2=a^2 + b^2 + 2ab\:\:\:\So,\\ \\ \implies (2x+1)^2 + (2x+3)^2 = 290 \\ \\ \implies [(2x)^2 + 1^2 + 2(2x)(1)] + [(2x)^2 + 3^2 + 2(2x)(3)] = 290 \\ \\ \implies 4x^2 + 1 + 4x + 4x^2 + 9 + 12x = 290 \\ \\ \implies 8x^2 + 16x + 10 = 290 \\ \\ \implies 2\!\!\!/(4x^2 + 8x + 5) = 290\!\!\!\!/^{145} \\ \\ \implies 4x^2 + 8x - 140 = 0 \\ \\ \implies 4(x^2 + 2x - 35) = 0 \\ \\ \implies x^2 + 2x - 35 = 0 \\ \\ \implies x^2 - 5x + 7x - 35 = 0 \\ \\ \implies x(x-5) + 7(x-5) = 0 \\ \\ \implies (x - 5)(x + 7) = 0 \\ \\ \implies x = 5 \& x = -7$

$\displaystyle The\:numbers\:are \\ 1.\: When\:x = 5 \\ \implies 2x + 1 => 2(5) + 1 => 10 + 1 => 11 \\ \implies 2x + 3 => 2(5) + 3 => 10 + 3 => 13 \\ \implies \bold{11 \: and\: 13} \\ 2.\: When\:x = -7 \\ \implies 2x + 1 => 2(-7) + 1 => -14 + 1 => -13 \\ \implies 2x + 3 => 2(-7) + 3 => -14 + 3 => -11 \\ \implies \bold{-11 \: and\: -13} \\ \bold{Hence\:the\: numbers\:are\:11\&13\:or\:-11\&-13.}$

Verification:

$\displaystyle When\:the\:numbers\:are\: 11\&13. \\ \implies (11)^2 + (13)^2 \\ \implies 121 + 169 \\ \implies 290 = RHS \\ When\:the\:numbers\:are\: -11\&-13. \\ \implies (-11)^2 + (-13)^2 \\ \implies 121 + 169 \\ \implies 290 = RHS$

Formula used:

• $(a+b)^2=a^2 + b^2 + 2ab$