Question

Find two natural number which is differ by 3 and the sum of whose squares is 117

Answer 1

x^2+(x+3)^2=117
x^2+x^2+6x+9=117
2x^2+6x-108=0
2.(x^2+3x-54)=0
2.(x-9).(x+6)=0
So x = -9 or x=6
The problem asked for natural numbers, so the smaller number is x=6 and the other number is 6+3=9

Answer 2

Given:–

• Sum of squares = 117
• The two natural numbers differs by 3

To find:–

• Two natural numbers

Concept:–

• Simple equations (based on quadratic equation)

Step by step explaination:–

Let us assume the first number be y. Therefore, second number will be (y+3).

According to the question.

⟼ y² + (y+3)² = 117

Now, opening brackets and solving.

⟼ y² + y² + 6y + 9 = 117

⟼ 2y² + 6x - 108 = 117

⟼ y² + 3y - 54 = 0

⟼ (y+9) (y-9) = 0

Therefore,

⟼ y = -9 and 6

Answer:–

6 is accepted as -9 is not a natural number and being negative it can't be accepted.

Second number = 6+3 ⟼ 9