Question

Divide 29 into two parts so that the sum of the squares of the parts is 425.

Answer 1

SOLUTION :  

GIVEN : Sum of the squares of two parts is 425.

Let x be the first part and second part be (29 - x).

A.T.Q  

⇒ x² + (29 - x)² = 425

⇒ x² + 841- 58x + x² = 425

[(a - b)² = a² + b² - 2ab]

⇒ x² + + x² - 58x + 841 -  425 = 0

⇒ 2x²  - 58x + 416 = 0

⇒ 2(x² - 29x + 208) = 0

⇒ x² - 29x + 208 = 0

⇒ x² - 16x - 13x + 208 = 0

⇒ x(x - 16) - 13(x - 16) = 0

⇒ (x - 16) (x - 13) = 0

⇒ (x - 16)  = 0  or  (x - 13) = 0

⇒ x = 13  or  x = 16

Case 1 : x = 13

When, first part is 13, then second part = (29 - x) =  29 - 13 = 16

Case 2 : x = 16

When, first part is 16, then second part = (29 - x) 29 - 16 = 13

Hence, the two parts are (13, 16) or (16, 13).

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Answer 2

Answer :

The two parts in which 29 can be divided so that the sum of the squares of the parts is 425 are (16,13) or (13,16).

Step-by-step explanation :

Given, the sum of the squares of two part is equal to 425.

Let the first part be x and second part be (29-x).

Then, according to question -

⇒ x² + (29 - x)² = 425

⇒ x² + 841- 58x + x² = 425 [$\because$(a - b)² = a² + b² - 2ab]

⇒ x² + + x² - 58x + 841 -  425 = 0

⇒ 2x²  - 58x + 416 = 0

⇒ 2(x² - 29x + 208) = 0

⇒ x² - 29x + 208 = 0

⇒ x² - 16x - 13x + 208 = 0

⇒ x(x - 16) - 13(x - 16) = 0

⇒ (x - 16)(x - 13) = 0

⇒ (x - 16) = 0 OR (x - 13) = 0

⇒ x = 13 OR x = 16

Now, considering them as individual cases -

• Case 1 -

If x = 13 then, second part = 16.

• Case 2 -

If x = 16 then, second part = 13