Answer :

Given: The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number.

To find: the numbers.

Solution:

Let one of the numbers be a.

Given, sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number.

2nd number = 2a – 3

According to given condition,

 a2 + (2a – 3)2 = 233

Apply the formula (x –y )2 = x2 + y2 -2xy on (2a – 3)2

a2 + 4a2 + 9 – 12a = 233

⇒ a2 + 4a2 + 9 – 12a - 233 = 0

5a2 – 12a – 224 = 0

5a2 – 40a + 28a – 224 = 0

5a(a – 8) + 28(a – 8) = 0

(5a + 28)(a – 8) = 0

⇒ (5a + 28) = 0  and (a – 8) = 0

⇒ a = -28/5 and a=8

To satisfy the given conditions a will be 8.

2nd number = 2(8) – 3 = 16-3 = 13

Thus the numbers are 8, 13.

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