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Find three consecutive odd integers such that the square of the first increased the product of the other two is 28.

Find Three Consecutive Odd Integers Such That The Square Of The First Increased The Product Of The Other Two Is 28 class=

Sagot :

Since these numbers are odds, they are of the form:

[tex]2x+1[/tex]

Let's take this as the first.

The second one must be:

[tex]2x+3[/tex]

And the last one:

[tex]2x+5[/tex]

The square of the first one increased the product of the other two is 28:

[tex](2x+1)^2+(2x+3)(2x+5)=28[/tex]

If we take k=2x+1, we get:

[tex]k^2+(k+2)(k+4)=28[/tex][tex]k^2+k^2+6k+8=28[/tex][tex]2k^2+6k+8=28[/tex][tex]2k^2+6k+8-28=0[/tex][tex]2k^2+6k-20=0[/tex][tex]k^2+3k-10=0[/tex][tex](k+5)(k-2)=0[/tex]

K could be -5 or 2, but because has to be a odd number k=-5.

Lets check:

[tex](-5)^2+(-3)(-1)[/tex][tex]25+3[/tex][tex]28[/tex]

The consecutive odd numbers are -5 , -3 and -1

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