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Ede4ka [16]
3 years ago
13

Find two positive consecutive odd integers such that square of the smaller integer is 10 more than the larger integer

Mathematics
1 answer:
Otrada [13]3 years ago
3 0

Let 2n+1 be the smaller integer. The larger integer is then 2n+3, and we have

(2n+1)^2=10+(2n+3)\implies4n^2+4n+1=2n+13

\implies4n^2+2n-12=0

\implies2n^2+n-6=0

\implies(2n-3)(n+2)=0

\implies 2n-3=0\text{ or }n+2=0

\implies n=\dfrac32\text{ or }n=-2

We omit n=-2, since 2(-2)+1=-3 is negative.

Then for n=\dfrac32 we find 2\left(\dfrac32\right)+1=4, but this is not odd.

There are no consecutive odd integers that satisfy the given condition!

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Step-by-step explanation:

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