Rozwiąż równanie kwadratowe"
F(X)=-x2+2x+3
delta= 2^2+4*3
delta=16
x=
f(x)=x2+3-10
f(x)=x2-7
delta=0^2+4*7
delta=28
f(x) = -x²+2x+3
Δ=b²-4ac=4+12=16 √Δ=4
x1=(-b-√Δ)/2a=(-2-4)/-2=-6/-2=3
x2=(-b+√Δ)/2a=(-2+4)/-2=2/-2=-1
odp.
x=-1 ∨ x=3
f(x)=x²+3-10=x²-7
x²-7=0
x=-√7 ∨ x=√7
lub jezeli miało być :
f(x)= x²+3x-10
Δ=9+40=49 √Δ=7
x1=(-3-7)/2=-10/2=-5
x2=(-3+7)/2=4/2=2
x=-5 ∨ x=2
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F(X)=-x2+2x+3
delta= 2^2+4*3
delta=16
x=
x=
f(x)=x2+3-10
f(x)=x2-7
delta=0^2+4*7
delta=28
x=
x=
f(x) = -x²+2x+3
Δ=b²-4ac=4+12=16 √Δ=4
x1=(-b-√Δ)/2a=(-2-4)/-2=-6/-2=3
x2=(-b+√Δ)/2a=(-2+4)/-2=2/-2=-1
odp.
x=-1 ∨ x=3
f(x)=x²+3-10=x²-7
x²-7=0
odp.
x=-√7 ∨ x=√7
lub jezeli miało być :
f(x)= x²+3x-10
Δ=9+40=49 √Δ=7
x1=(-3-7)/2=-10/2=-5
x2=(-3+7)/2=4/2=2
odp.
x=-5 ∨ x=2