suppose 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
How do you find quadratic equations if you are only gven the solution?
5 answers
Is it possible to have diferent quadratic equations with the same solution? My instinct says yes, my fear of math says no.
no
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
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
change
"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
"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
Wow! 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?