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Three Consecutive Natural Numbers Are Such that the Square of the Middle Number Exceeds the Difference of the Squares of the Other Two by 60. Assume the Middle Number to Be X and Form a - Mathematics

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Question

Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60.
Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.

Solution

Let the numbers be x – 1, x and x + 1.
From the given information,
x2 = (x + 1)2 – (x – 1)2 + 60
x2 = x2 + 1 + 2x – x2 – 1 + 2x + 60
x2 = 4x + 60
x2 – 4x – 60 = 0
(x – 10) (x + 6) = 0
x = 10, -6
Since, x is a natural number, so x = 10.
Thus, the three numbers are 9, 10 and 11.

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 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Exercise 6(A) | Q: 13 | Page no. 70
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Three Consecutive Natural Numbers Are Such that the Square of the Middle Number Exceeds the Difference of the Squares of the Other Two by 60. Assume the Middle Number to Be X and Form a Concept: Quadratic Equations.
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