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Question

Three consecutive positive integers are taken such that sum of square of the first number and the product of other two numbers is 232.The numbers are.

A
11, 12, 13
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B
10, 11, 12
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C
7, 8, 9
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Solution

The correct option is B 10, 11, 12
Let the three consecutive positive integers be x, x+1 and x+2.

Given that, sum of square of the first number and the product of other two numbers is 232
x2+(x+1)(x+2)=232
x2+(x2+3x+2)=232
2x2+3x230=0
Using quadratic formula, x=b±b24ac2a

Here a=2,b=3 and c=230

x=3±324×2×2304
=3±18494
=3±434
x=10 or 11.5
Ignoring negative value, since x is a positive integer.
x = 10.
Hence, the numbers are 10, 11 and 12.

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