Question

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

Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Assumption,

First number be p

Also,

Second number be p - 3

Situation,

(p)² + (p - 3)² = 117

p² + p² + 9 - 6p = 117

2p² + 9 - 6p - 117 = 0

2p² - 6p - 108 = 0

p² - 3p - 54 = 0

Splitting middle term,

p² - 3p - 54 = 0

p² - 9p + 6p - 54 = 0

p(p - 9) + 6(p - 9) = 0

(p + 6)(p - 9) = 0

p = 9 and p = -6

Here,

p should be positive.

Hence,

p = 9

Also,

p - 3 = 6

Therefore,

Two numbers be 9 and 6.

Answer 2

Answer

Given,

• Two natural numbers differ by 3 and their squares have the sum 117

To find,

• The two natural numbers

Solution ,

Let the First number be : x

Let the second number be : y

According to the question :

y = x - 3

Also,

⇛ $\rm x^2 + (x - 3)^2 = 117$

⇛ $\rm x^2 + x^2 + 9 - 6x = 117$

⇛ $\rm 2x^2 + 9 - 6x - 117 = 0$

⇛ $\rm 2x^2 - 6x - 108 = 0$

Dividing the equation by 2

⇛ $\rm x^2 - 3x - 54 = 0$

Here we got a quadratic equation .Let us solve it by Splitting middle term method.

⇛ $\rm x^2 - 3x - 54 = 0$

⇛ $\rm x^2 - 9x + 6x - 54 = 0$

⇛ $\rm x(x - 9) + 6(x - 9) = 0$

⇛ $\rm (x + 6)(x - 9) = 0$

• x = 9
• x = -6

Here,

'x' need to be positive.

Hence,

x = 9

Also,

x - 3 = y

So ,

y = 6

Hence two natural numbers will be 9 & 6