Question

The sum of square of 2 consecutive odd natural numbers is394. Then the value of smaller number is?
pls answer fast!!

Answer 1

Answer:

Given :-

• The sum square of 2 consecutive odd natural numbers is 394.

To Find :-

• What is the smaller number.

Solution :-

Let,

➦ First Number = x

➦ Second Number = x + 2

According to the question,

↦ x² + (x + 2)² = 394

↦ x² + x² + 2(x)(2) + (2)² = 394 [ (a + b)² = a² + 2ab + b² ]

↦ x² + x² + 4x + 4 = 394

↦ 2x² + 4x + 4 - 394 = 0

↦ 2x² + 4x - 390 = 0

↦ 2(x² + 2x - 195) = 0

↦ x² + 2x - 195 = 0 × 2

↦ x² + 2x - 195 = 0

↦ x² + (15 - 13)x - 195 = 0

↦ x² + 15x - 13x - 195 = 0

↦ x(x + 15) - 13(x + 15) = 9

↦ (x + 15)(x - 13) = 0

↦ (x + 15) = 0

↦ x + 15 = 0

➠ x = - 15

↦ (x - 13) = 0

↦ x - 13 = 0

➠ x = 13

We can't take x as negative (- ve)

So, the x is 13.

Hence, the required numbers are :

❒ First Number :

⇒ First Number = x

➲ First Number = 13

❒ Second Number :

⇒ Second Number = x + 2

⇒ Second Number = 13 + 2

➲ Second Number = 15

∴ The smaller number is 13.

Answer 2

Step-by-step explanation:

$\underline{\purple{\ddot{\Mathsdude}}}$

♧♧Given:-

• The sum of square of 2 consecutive odd natural number is 394.

♧♧To prove:-

• The value of smaller number

♧♧Proof:-

♡♡Lets take ,

• First number should be x
• Second number should be x+2

♧♧From the question ,

• x^2+(x+2)^2=394.

• x^2+x^2+2 (x)(2)+(2)^2=394

●(a+b)^2=a^2+b^2+2ab)

• x^2+x^2+4x+4=394

• 2x^2+4x+4-394=0

• 2x^2+4x+390=0

• 2 (x^2+2x+2-195)=0

• x^2+2x-195=0×2

• x^2+2x-195=0

• x^2+15x-13x-195=0

• x (x+15)-13 (x+15)=0

• x+15=0 x-13=0
• x=-15 x=+13

♧♧We will not take x as negative so ,

• The number is 13.

♧♧According to the question,

1. First number=13
2. Second number =x+2

=13+2=15.

♧♧Therefore, the smaller number is 13.

♧♧Hope it helps u mate .

♧♧Thank you .