ad 1
4(x-3)(x+1)=0
(4x-12)(x+1)=0
4x-12=0 V x+1=0
x=3 V x=-1
Odp : D
ad2
a)
-5x²+2x=0
Δ= b²-4ac
Δ= 4- 4 * (-5) *0
Δ=4
√Δ=2
x1 = -b-√Δ /2a = -2-2/2*(-5) = -4/-10 = 2/5
x2 = -b+√Δ /2a = -2+2/10 = 0/10 = 0
b) 6x² = 24x
6x²-24x = 0
6x( x -4) =0
6x=0 V x-4=0
x=0 V x=4
c) x(x-3)=9x(x-3)
x²-3x= 9x²-27x
x²-3x-9x²+27x = 0
-8x²+24x =0
-8x(x-3) =0
-8x=0 V x-3 = 0
x=0 V x=3
ad 3
-4x² +25x -6≥ 0
Δ= b²-4ac
Δ=529
√Δ=23
x1 = 6
x2= 1/4
oś i parabola ramionami w dół,
x∈<1/4 ; 6>
ad 4
f(x) = [tex]\sqrt{-x^2+4x+5}[/tex]
-x²+4x+5≥0
Δ=b²-4ac
Δ=36
√Δ=6
x1=5
x2=-1
os i parabola ramionami do dołu ,
x∈<-1;5>
odp; D
ad 5
2x²-3x-20 < 0
Δ=b²-4ac
Δ=169
√Δ=13
x1= 2,5
x2=4
parabola ramionami do góry,
x∈(2,5 ; 4)
