Question

find two natural number which is which differ by 3 and whose squares have the sum 117​

Answer 1

Step-by-step explanation:

Hi Dear !!

Let the two natural number be X and y .

A/Q,

X - Y = 3 -----------(1)

x² + y² = 117 --------(2)

From equation (1) , we get

x - y = 3

x = ( 3 + y ) ------(3)

Putting the Value of x in equation (2) , we get

x² - y² = 117

( 3 + y )² - y² = 117

3² + y² + 2 * 3 *y - y² =117[Since,(a+b)²=a²+b²+2ab]

9 + y² + 6y - y² = 117

6y = 117 - 9

6y = 108

y = 108/6 = 18

Y = 18

Putting the value of Y in equation (3), we get

x = 3 + y

x = 3 + 18 = 21.

Hence,

X = 21 and Y = 18

Answer 2

Answer:

• Two numbers be 9 and 6.

Step-by-step explanation:

Given:

• Two natural number which is differ by 3 and whose squares have the sum 117.

To Find:

• The number.

First number be 'x'

And, second number be 'x - 3'.

Now, according to question,

⇒ (x)² + (x - 3)² = 117

⇒ x² + x² + 9 - 6x = 117

⇒ 2x² + 9 - 6x - 117 = 0

⇒ 2x² - 6x - 108 = 0

⇒ x² - 3x - 54 = 0

Now, we will solve this equation by splitting middle term,

⇒ x² - 3x - 54 = 0

⇒ x² - 9x + 6x - 54 = 0

⇒ x(x - 9) + 6(x - 9) = 0

⇒ (x + 6) (x - 9) = 0

⇒ x = 9 and -6.

Hence, x should be positive. So x = 9

And, x - 3 = 6

Hence, Two numbers be 9 and 6.