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February 1, 2015

February 1, 2015

Posted by **Marysvoice** on Sunday, February 3, 2008 at 4:42pm.

- Algebra -
**Reiny**, Sunday, February 3, 2008 at 5:13pmsuppose you know the solutions are 5 and -3

form factors (x-5) and (x+3), multiply them and set the result equal to zero

(x-5)(x+3)=0

x^2 -2x -15=0

- Algebra -
**Marysvoice**, Sunday, February 3, 2008 at 5:17pmIs it possible to have diferent quadratic equations with the same solution? My instinct says yes, my fear of math says no.

- Algebra -
- Algebra -
**Reiny**, Sunday, February 3, 2008 at 5:27pmno

any quadratic equation, which has been simplied to its simplest form has a unique solution, and given an set of solutions there is only one unique quadratic equation

If I take the equation from my example and multiply it by 3 I would get

3x^2 - 6x - 45 = 0

it really is not a "different" equation.

BTW, if your solutions had been fractions, say 4/5 and -2/3, your two factors would have been

(5x-4) and (3x+2)

There is another way:

If you know the two solutions are m and n, then you can just form the quadratic by writing

x^2 - (m+n)x + mn = 0, then simplify

eg. going back to my example of solutions of 5 and -3, their sum is 2 and their product is -15

the equation would be x^2 -(2)x + (15) = 0 , just like above

- Algebra -
**Reiny**, Sunday, February 3, 2008 at 5:29pmchange

"the equation would be x^2 -(2)x + (15) = 0 , just like above" to

the equation would be x^2 -(2)x + (-15) = 0 , just like above

- Algebra -
**Marysvoice**, Sunday, February 3, 2008 at 5:33pmWow! I have a lot to learn. I don't mean to take up your time, but I understand there is more than one way to solve quadratic equations. Is there a way you think is easier and how?

- Algebra -

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