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# (i) Find three successive even natural numbers, the sum of whose squares is 308. (ii) Find three consecutive odd integers, the sum of whose squares is 83.

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This question has been taken from Book:- ML aggarwal, Avichal publication, class10th, quadratic equation in one variable, chapter 5, exercise 5.5
This is an important ques

(i) Find three successive even natural numbers, the sum of whose squares is 308. (ii) Find three consecutive odd integers, the sum of whose squares is 83.
Question no.8 , ML Aggarwal, chapter 5, exercise 5.5, quadratic equation in one variable, ICSE board,

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1. Solution:

(i) Find three successive even natural numbers, the sum of whose squares is 308.

Let us consider first even natural number be ‘2x’

Second even number be ‘2x + 2’

Third even number be ‘2x + 4’

So according to the question,

(2x)2 + (2x + 2)2 + (2x + 4)2 = 308

4x2 + 4x2 + 8x + 4 + 4x2 + 16x + 16 – 308 = 0

12x2 + 24x – 288 = 0

Divide by 12, we get

x2 + 2x – 24 = 0

Let us factorize,

x2 + 6x – 4x – 24 = 0

x(x + 6) – 4(x + 6) = 0

(x + 6) (x – 4) = 0

So,

(x + 6) = 0 or (x – 4) = 0

x = -6 or x = 4

∴ Value of x = 4 [since, -6 is not positive]

First even natural number = 2x = 2 (4) = 8

Second even natural number = 2x + 2 = 2(4) + 2 = 10

Third even natural number = 2x + 4 = 2(4) + 4 = 12

∴ The numbers are 8, 10, 12.

(ii) Find three consecutive odd integers, the sum of whose squares is 83.

Let the three numbers be ‘x’, ‘x + 2’, ‘x + 4’

So according to the question,

(x)2 + (x + 2)2 + (x + 4)2 = 83

x2 + x2 + 4x + 4 + x2 + 8x + 16 – 83 = 0

3x2 + 12x – 63 = 0

Divide by 3, we get

x2 + 4x – 21 = 0

let us factorize,

x2 + 7x – 3x – 21 = 0

x(x + 7) – 3 (x + 7) = 0

(x + 7) (x – 3) = 0

So,

(x + 7) = 0 or (x – 3) = 0

x = -7 or x = 3

∴ The numbers will be x, x+2, x+4 => -7, -7+2, -7+4 => -7, -5, -3

Or the numbers will be x, x+2, x+4 => 3, 3+2, 3+4, => 3, 5, 7

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