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Three consecutive positive integer are such …

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Three consecutive positive integer are such that the sum of the square of the first and product of the other two is 46 find the integers
  • 1 answers

Sia ? 6 years, 5 months ago

Let the integers  be x, x + 1 and x + 2. Then,
x2 + (x + 1)(x + 2) = 46
{tex}\Rightarrow{/tex} x2 + (x2 + 2x + 1x + 2) = 46
{tex}\Rightarrow{/tex} 2x2 + 3x + 2 = 46
{tex}\Rightarrow{/tex} 2x2 + 3x + 2 - 46 = 0
{tex}\Rightarrow{/tex} 2x2 + 3x - 44 = 0
{tex}\Rightarrow{/tex} 2x2 + 11x - 8x - 44 = 0
{tex}\Rightarrow{/tex} x(2x + 11) - 4(2x + 11) = 0
{tex}\Rightarrow{/tex} (x - 4)(2x + 11) = 0
{tex}\Rightarrow{/tex} x - 4 = 0 [{tex}\because{/tex}  x  is an integer {tex}\therefore{/tex} 2x + 11 {tex}\ne{/tex} 0]
{tex}\Rightarrow{/tex} x = 4
{tex}\therefore{/tex} x + 1 = 4 + 1 = 5
And, x + 2 = 4 + 2 = 6
Hence, required numbers are 4, 5 and 6.

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