The difference of two natural number is 3 and their square have the sum 117
Answers
Answered by
0
X, Y
X-Y=3
X = 3+Y
X²+ Y² =117
(3+Y)² +Y² = 117
3²+Y²+6Y+Y² =117
9+ 2Y² + 6Y =117
then solve
X-Y=3
X = 3+Y
X²+ Y² =117
(3+Y)² +Y² = 117
3²+Y²+6Y+Y² =117
9+ 2Y² + 6Y =117
then solve
Answered by
30
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
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