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Show that if x1 and x2 are real solutions of x2+bx+c=0, then x1+x2=-b and x1x2=c.

(In the equation x2+bx..., 2 is squared for x. All the other x1 and x2 are just the names of the x's).

Great if somebody could crack this nut!

jr.heiberg wrote:Show that if x1 and x2 are real solutions of x2+bx+c=0, then x1+x2=-b and x1x2=c.

(In the equation x2+bx..., 2 is squared for x. All the other x1 and x2 are just the names of the x's).

Great if somebody could crack this nut!

I am not 100% confident about this, but:

Show that if [tex]x_1[/tex] and [tex]x_2[/tex] are real solutions of [tex]x^2 +bx +c = 0[/tex]
Then [tex]x_1 + x_2 = -b[/tex]
And [tex]x_1 \cdot x_2 = c[/tex]

In that case;

[tex](x+x_1)(x+x_2) = 0[/tex]

As you can see, the product of [tex]x_1[/tex] and [tex]x_2[/tex] equals [tex]c[/tex]

In addition, we can set x=1, factor it out like this:
[tex](1+x_1)(1+x_2) = 1 + (x_2 +x_1) + x_1x_2 [/tex]

If we only take the expression within the parenthesis into consideration, we have: [tex]x_2+x_1[/tex] which is b.

Was this solution satisfying enough, or would you like me to elaborate on it?

for det første må polynomet skrives slik: [tex](x-x_1)(x-x_2)[/tex], for det andre trenger du ikke i det hele tatt å anta at et polynom kan skrives som lineære faktorer hvor hver har et nullpunkt hvor polynomet har et nullpunkt.

Prøv å løse annengradslikningen [tex]x^2+bx+x=0[/tex], og bruk disse konkrete løsningene til å bevise det om skal bevises.

EDIT: han her er kanskje engelsk ja

To jr.heiberg:

Find the solutions of [tex]x^2+bx+c=0[/tex] in terms of b and c, and try to prove it by substituting them for [tex]x_1[/tex] and [tex]x_2[/tex].

If [tex]x_1,x_2[/tex] are two solutions of [tex]x^2 + bx + c = 0[/tex] then it is known that the polynomial can be factorized into [tex](x-x_1)(x-x_2)[/tex]. Expand and compare.

Jarle10 wrote:for det første må polynomet skrives slik: [tex](x-x_1)(x-x_2)[/tex], for det andre trenger du ikke i det hele tatt å anta at et polynom kan skrives som lineære faktorer hvor hver har et nullpunkt hvor polynomet har et nullpunkt.

Prøv å løse annengradslikningen [tex]x^2+bx+x=0[/tex], og bruk disse konkrete løsningene til å bevise det om skal bevises.

EDIT: han her er kanskje engelsk ja

To jr.heiberg:

Find the solutions of [tex]x^2+bx+c=0[/tex] in terms of b and c, and try to prove it by substituting them for [tex]x_1[/tex] and [tex]x_2[/tex].

Hva kan jeg si? Man lærer visst av å tro man kan også, hehehe

Hva kan jeg si? Man lærer visst av å tro man kan også, hehehe

Beste i år

I would use the factored form of a second degree polynominal [tex](x-x_1)(x-x_2)[/tex] as a starting point, expand it and compare my result with [tex]x^2+bx+c[/tex].

It shouldn't be a problem for anyone familiar with basic algebra.