Respuesta :
Answer:
i) -2
Step-by-step explanation:
In mathematics, Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.
c and d are solutions(roots) of [tex]x^2 + Ax + B=0[/tex]
a and b are solutions(roots) of [tex]x^2 + Cx + D = 0[/tex]
[tex]c+d=-a[/tex] eq. 1
[tex]cd=b[/tex] eq. 2
[tex]a+b=-c[/tex] eq. 3
[tex]ab=d[/tex] eq. 4
from eq. 1 we get [tex]d=-a-c[/tex]
from eq. 3 we get [tex]b=-a-c[/tex]
so [tex]d=b[/tex]
now substitute [tex]d=b[/tex] in eq. 2
[tex]cb=b[/tex]
[tex]c=b/b=1[/tex]
now substitute [tex]d=b[/tex] in eq. 4
[tex]ab=d[/tex]
[tex]ad=d[/tex]
[tex]a=d/d=1[/tex]
now substitute the values of [tex]c=1 and a=1[/tex] in eq. 1
[tex]d=-a-c=-1-1=-2[/tex]
similarly, substitute the values of [tex]c=1 and a=1[/tex] in eq. 3
[tex]b=-a-c=-1-1=-2[/tex]
Finally,
[tex]A+B+C+D= 1-2+1-2=-2[/tex]
Lets verify and see if our answer is right!
[tex]x^2 + Ax + B =0[/tex]
Substituting the values of A and B
[tex]X^2+x-2=0[/tex]
we know that c and d are solutions of this equation so they must satisfy the equation.
put [tex]c=1[/tex]
[tex](1)^2+1-2=0[/tex]
[tex]2-2=0[/tex]
[tex]0=0[/tex] (proved)
put [tex]d=-2[/tex]
[tex](-2)^2-2-2=0[/tex]
[tex]4-4=0[/tex]
[tex]0=0[/tex] (proved)
