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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
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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.
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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