Wednesday

October 1, 2014

October 1, 2014

Posted by **Gabby** on Saturday, March 15, 2014 at 3:57pm.

- Algebra Question~! -
**Steve**, Saturday, March 15, 2014 at 4:56pm(x+10)(x-3)

- Algebra Question~! -
**Reiny**, Saturday, March 15, 2014 at 5:00pmx^2 + 7x - 30

= (x+10)(x-3)

so the sides could be (x+10) and (x-3) , were x > 3

or

x^2 + 7x - 30

= (x-2)(x+9 - 12/(x-2) )

so the sides could be x-2 and x+9 - 12/(x-2)

or

x^2 + 7x - 30

= (x+5)((x+2 - 40/(x+5) )

so the sides could be x+5 and x+2 - 40/(x+5)

or ...

as you can see there is no unique solution, I could pick any binomial for a first side, do a division and get the 2nd side.

Illustrate with numbers

Suppose you have an area of 30

so the sides could be 5 and 6

or

the sides could be 3 and 10

or

the sides could be 1 and 30

or

4.5 and 6 2/3

etc, can you see my point?

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