Question

my text says that the roots of an quadratic equation when added together are equal to -b/a and c/a when multiplied. is it possible to prove this?

Answer 1

i guess you can prove this from formula for the roots of the equation

x = [-b + sqrt(b^2 - 4ac) ]/ 2a or [-b - sqrt(b^2 - 4ac) ]/ 2a

adding these 2 expressions should simplify to -b/a

there might be a simpler way to do this though....

Answer 2

yes. add and multiply the roots \[r _{1}=\frac{ -b }{ 2a }+\frac{ \sqrt{b^2-4ac} }{ 2a }\] and \[r _{2}=\frac{ -b }{ 2a }-\frac{ \sqrt{b^2-4ac} }{ 2a }\]

Answer 3

If D>0 then there are two distinct roots given by\[\LARGE{\alpha = \frac{-b + \sqrt{b^2-4ac}}{2a}}\]\[\LARGE{\frac{-b-\sqrt{b^2-4ac}}{2a}}\]
If D=0 then the roots are real & equal \(\Large{\alpha + \beta=\frac{-b}{a}}\).

If D<0 there are no real roots
sum of the roots \(\Large{\alpha+\beta=\frac{-b}{a}}\)
and the products of the roots \(\Large{\alpha\beta=c/a}\)
Therefore the quadratic equation is \(\Large{x^2-(\alpha + \beta)x+(\alpha\beta)=0}\)

Answer 4

using this tips u can prove :)

Answer 5

cool i shall give it a try

Answer 6

sure :)