Can anyone explain how to write a quadratic equation (ax^2 + bx+c=0) with a given real roots??

for example


c) 3+ root 2, 3 - root 2





root=the root sign

Can anyone explain how to write a quadratic equation (ax^2 + bx+c=0) with a given real roots??
Subtract each root from x, and then multiply


(x-(3+sqrt(2))) (x-(3-sqrt(2)))


FOIL


= x^2 - x(3-sqrt(2)) - x(3+sqrt(2)) + (3+sqrt(2))(3-sqrt(2))


= x^2 - 3x + x*sqrt(2) - 3x - x*sqrt(2) + 9 - 3sqrt(2) + 3sqrt(2)


- sqrt(2)*sqrt(2)


= x^2 - 6x + 9 - 2


= x^2 - 6x + 7
Reply:(x - (3 +root2))(x - (3 - root2))=0





Then multiply it out.
Reply:Some review: we know that a root of a polynomial, let's say r, is such that that P(r) = 0. That is if we have a P(x) = x^2 + x - 3, and a root r, P(r) = r^2 + r - 3 = 0. So a quadractic equation is just a special case of a polynomial and we know that they have at most 2 roots. From your example, r = 3+\sqrt{2}, t = 3-\sqrt{2}, r,t are roots of the polynomial, thus P(r) = 0 and P(t) = 0. Now the problem becomes simple, if we let P(x) = (r-x)(t-x), it is easy to see that P(r) = 0 and P(t) = 0. Thus, P(x) = (3+\sqrt{2} - x)(3-\sqrt{2}-x), multiply them to check that they are indeed quadratic equations and you are done!