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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

User Crowne
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1 Answer

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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
User Sogu
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