Question

Se considera x1 si x2 solutiile ecuatiei x²-4x+m=0, unde m este numar real. Determinati numarul real m pentru care x1²+x2²=2

Answer 1

x1,x2€f(x)

f(x)=x^2-4x+m=0

a=1

b=-4

c=m

Delta: b^2-4ac=16-4m

x1=(4-√16-4m/2a=4-4-√4m)/2a

x2=(4+√16-4m/2a=4+4-√4)/2a

x1=-√4m/2a|n^2

x1=-4m/4

x2=8-√4m/2|n^2

x2=(64-32√m+4m)/4

(-4m/4+64-32√m+4m)/4=2

(-4m+64-32√m+4m)/4=2|*4

64-32√m=8

-32√m=-56

√m=-56/-32

√m=1.75

m=1.75^2

m=3.0625