The sum of the squares of two consecutive natural numbers is 365

Q) The sum of the squares of two consecutive natural numbers is 365. Find the numbers.

Ans:

Step 1: Let the first number be X.

Since the numbers are consecutive, hence next number will be: X + 1

Step 2: According to the given condition, sum of the squares of these two consecutive numbers is 365:

∴ (X)² + (X + 1) ²= 365

∴ X ² + X ² + 2 X + 1 = 365

∴ 2 X ² + 2 X + 1 = 365

∴ 2 X ² + 2 X – 364 = 0

Dividing by 2, we get:

X ² + X – 182 = 0

Step 3: Solving for X:

Method 1: Solving using the quadratic formula:

By comparing with standard quadratic equation, a x ²+ b x + c = 0, we have a = 1, b = 1, c = – 182

Now, we put values in the quadratic formula:

∴ X = $\frac{- (1) \pm \sqrt{(1)^2 - 4(1)(-182)}}{2(1)}$

∴ X = $\frac{- 1 \pm \sqrt{1 + 728}}{2}$

∴ X = $\frac{- 1 \pm \sqrt{729}}{2}$

∴ X = $\frac{-1 \pm 27}{2}$

Solving for positive X:

X = $\frac{- 1 + 27 }{2}$ = $\frac{26}{2}$ = 13

and X + 1 = 13 + 1 = 14

(Here, we reject X = -14, because we need to take only natural numbers)

Therefore, the two consecutive numbers are: 13 and 14.

Method 2: By mid-term splitting, we get:

X ² + X – 182 = 0

∴ X ² + 14 X – 13 x – 182 = 0 (182 = 14 x 13)

∴ X (X + 14) – 13 ( X + 14) = 0

∴ (X – 13) (X + 14) = 0

∴ X = 13 and – 14

(Here, we reject X = – 14, because we need to take only natural numbers)

∴ X = 13 and X + 1 = 13 + 1 = 14

Therefore, the two consecutive numbers are: 13 and 14.

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