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Grace [21]
3 years ago
11

The product of two consecutive even integers is 288 find the integers.

Mathematics
1 answer:
algol [13]3 years ago
5 0
Using algebra
n and n+2 are the integers
n*(n+2)=288
n^2+2n=288
n^2+2n-288=0
288=2*2*2*2*2*3*3=(2*2*2*2)*(2*3*3)=16*18 and 18-16=2
factor(n+18)(n-16)=0
n+18=0
n=-18
n+2=-16
this is one solution
n-16=0
n=16
n+2=18
this is another solution
as you can see, it's just a matter of factoring 288using arithmetic...√288=16.97≈1716 and 18 are the integers (as well as -16 and -18)
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9514 1404 393

Answer:

  D.

Step-by-step explanation:

The wording "when x is an appropriate value" is irrelevant to this question. That phrase should be ignored. (You may want to report this to your teacher.)

When you look at the answer choices, you see that all of them are negative except the last one (D). When you look at the problem fraction, you see that it is positive.

The only reasonable choice is D.

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Your calculator can check this for you.

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If you want to "rationalize the denominator", then multiply numerator and denominator by the conjugate of the denominator. The conjugate is formed by switching the sign between terms.

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<em>Additional comment</em>

We "rationalize the denominator" in this way to take advantage of the relation ...

  (a -b)(a +b) = a² -b²

Using this gets rid of the irrational root in the denominator, hence "rationalizes" the denominator.

We could also have multiplied by (3 -√3)/(3 -√3). This would have made the denominator positive, instead of negative. However, I chose to use (√3 -3) so you could see that all we did was change the sign from (√3 +3).

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