Question

Find two consecative odd positive integers sum of whose squares is 290? (hint x and x+2) consecative odd+ve integer​

Answer 1

SOLUTION:-

Given

• Sum of Squares = 290

To find

• Two consecutive odd positive integers

Explanation

Let numbers be x and x+2

Now, it is Given sum of Square = 290

➡️ x² + (x+2)² = 290

➡️ x² + x²+2² + 2×x× 2= 290

➡️ x² + x²+ 4+ 4x= 290

➡️ 2x²+4x+4=290

➡️ 2x²+4x+4-290=0

➡️ 2x²+4x-286=0

Dividing by 2, we get ⤵️

➡️ x²+2x- 143=0

➡️ x²+ (13-11)x-143=0

➡️ x²+13x-11x-143=0

➡️ x(x+13)-11(x+13)=0

➡️ (x-11)(x+13)=0

➡️ x=11, x=-13

In the question we are given with positive integers.

Hence , the number is 11 and consecutive odd number is 13.

Thus, These are required numbers.

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Answer 2

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• Sum of whose square two odd positive integers = 290

$\Large{\underline{\mathfrak{\orange{\bf{Find}}}}}$

• These two consecutive odd positive integers

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

Let,

• First odd positive integers = x
• Second odd positive integers = (x + 2)

A/C to question,

(Sum of whose square two odd positive integers = 290 )

➠ (x)² + (x +2)² = 290

➠ (x² + x² + 2x + 4 ) = 290

➠ 2x² + 2x + 4 - 290 = 0

➠ 2x² + 2x - 286 = 0

➠ x² + x - 143 = 0

➠ x² + 13x - 11x - 143 = 0

➠ x(x+13) - 11(x+13) = 0

➠ (x+13)(x-11) = 0

➠(x+13) = 0. (x - 11) = 0

➠ x = - 13. Or, x = 11

According to question,

Given consecutive odd number is positive .

So, All Value of x be always positive

$\Large{\underline{\underline{\mathfrak{\red{\bf{Hence}}}}}}$

• Value of x be = 13 & 11.

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