The product of two consecutive integers is 98 more than the next integer. What is the largest of the three integers?

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The product of two consecutive integers is 98 more than the next integer. What is the largest of the three integers?

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Answer 1 · 2016-07-31T19:26:36

So the three integers are $10$ $10$ , $11$ $11$ , $12$ $12$

Explanation:

Let $3$ $3$ consecutive integers be $(a - 1), a and (a + 1)$ $(a-1),a and (a+1)$
Therefore
$a (a - 1) = (a + 1) + 98$ $a(a-1)=(a+1)+98$
or
$a^{2} - a = a + 99$ $a^2-a=a+99$
or
$a^{2} - 2 a - 99 = 0$ $a^2-2a-99=0$
or
$a^{2} - 11 a + 9 a - 99 = 0$ $a^2-11a+9a-99=0$
or
$a (a - 11) + 9 (a - 11) = 0$ $a(a-11)+9(a-11)=0$
or
$(a - 11) (a + 9) = 0$ $(a-11)(a+9)=0$
or
$a - 11 = 0$ $a-11=0$
or
$a = 11$ $a=11$
$a + 9 = 0$ $a+9=0$
or
$a = - 9$ $a=-9$
We will take only positive value So $a = 11$ $a=11$
So the three integers are $10$ $10$ , $11$ $11$ , $12$ $12$

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The product of two consecutive integers is 98 more than the next integer. What is the largest of the three integers?

1 Answer

Explanation:

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