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 $10$ , $11$ , $12$

Explanation:

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

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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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