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3 May, 10:29

Find three consecutive multiples of 5 so that the square of the third number, decreased by 5 times the second number is the same as 25 more than twice the product of the first two numbers

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  1. 3 May, 11:54
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    Let the unknown value be 'x'

    Consider the three consecutive multiples of 5 as 'x', ' (x + 5) ' and ' (x + 10) '

    First number = x

    Second number = (x + 5)

    Third number = (x + 10)

    Using the above statement, "The square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers," let us form an equation

    (x + 10) ² - 5 (x + 5) = 25 + (2 (x) (x + 5))

    x² + 20x + 100 - 5x - 25 = 25 + 2x² + 10x

    x² + 15x + 75 = 2x² + 10x + 25

    Bringing everything to one side,

    0 = 2x² + 10x + 25 - x² - 15x - 75

    or

    2x² + 10x + 25 - x² - 15x - 75 = 0

    x² - 5x - 50 = 0

    Factorising the above equation,

    (x - 10) (x + 5) = 0 ⇒ x = 10 or x = - 5

    If x = 10, then the three consecutive numbers are 10, 15, 20.

    Let us check if it follows the above statement!

    Square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers

    20² - (5 x 15) = 25 + (2 x (10 x 15))

    400 - 75 = 25 + (2 x 150)

    325 = 25 + 300

    325 = 325

    Now let us try with x = - 5. The three consecutive numbers are - 5, 0, 5

    Let us check if it follows the above statement.

    Square of the third number, decreased by 5 times the second number = 25 more than twice the product of the first two numbers

    5² - (5 x 0) = 25 + (2 x (-5 x 0))

    25 - 0 = 25 + (2 x 0)

    25 = 25 + 0

    25 = 25

    Hence both x = 10 and x = - 5 are correct
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