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

Answer 1

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

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#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Jace Anderson

Let's denote the first integer as ( x ) and the next consecutive integer as ( x + 1 ). According to the given information, the product of these two consecutive integers is ( 98 ) more than the next integer, which can be expressed as:

[ x(x + 1) = (x + 1) + 98 ]

Solving this equation:

[ x^2 + x = x + 99 ] [ x^2 = 99 ] [ x = \sqrt{99} ]

Since we're looking for integers, the largest integer that is less than or equal to ( \sqrt{99} ) is ( 9 ). So, the largest of the three integers is ( x + 1 = 9 + 1 = 10 ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.