Question

Rozwiąż podane równania. (rów. kwadratowe)

Ile zrobisz będzie ok.

Answer 1

Wzory:

Δ=b²-4ac

x₁=[-b-√Δ]/2a

x₂=[-b+√Δ]/2a

Wzory skróconego mnożenia:

(a+b)²=a²+2ab+b² - kwadrat sumy;

(a-b)²=a²-2ab+b² - kwadrat różnicy;

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a) x²+4x-12=0

Δ=4²-4*1*(-12)=16+48=64

√Δ=8

x₁=[-4-8]/2=-6

x₂=[-4+8]/2=2

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b) 2x²+5x-3=0

Δ=5²-4*2*(-3)=25+24=49

√Δ=7

x₁=[-5-7]/4=-3

x₂=[-5+7]/4=1/2

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c) x²-x-1=0

Δ=(-1)²-4*1*(-1)=1+4=5

√Δ=√5

x₁=[1-√5]/2

x₂=[1+√5]/2

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d) 4x²+4x+1=0

(2x+1)²=0

2x+1=0

x=-1/2

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e) x²-6x-1=0

Δ=(-6)²-4*1*(-1)=36+4=40

√Δ=2√10

x₁=[6-2√10]/2=3-√10

x₂=[6+2√10]/2=3+√10

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f) 1/3 x² -4x+12=0

Δ=(-4)²-4* 1/3 *12=16-16=0

x=-b/2a

x=4/(2/3)

x=4 * 3/2

x=6

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g) -x²+2x+4=0

Δ=2²-4*(-1)*4=4+16=20

√Δ=2√5

x₁=[-2-2√5]/(-2)=2+√5

x₂=[-2+2√5]/(-2)=2-√5

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h) 1/4 x²-5x+25=0

(1/2 x - 5)²=0

1/2 x - 5=0

x=10

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i) 5x²+2x-1=0

Δ=2²-4*5*(-1)=4+29=24

√Δ=2√6

x₁=[-2-2√6]/10=-[1+√6]/5

x₂=[-2+2√6]/10=-[1-√6]/5

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j) x²+2x+9=0

Δ=2²-4*1*9=4-36=-32

Δ<0 - brak rozwiązań

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k) -x²+2√2x-1=0

Δ=(2√2)²-4*(-1)*(-1)=8-4=4

√Δ=2

x₁=[-2√2-2]/(-2)=√2+1

x₂=[-2√2+2]/(-2)=√2-1

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l) x²-2√15x-1=0

Δ=(-2√15)²-4*1*(-1)=60+4=64

√Δ=8

x₁=[2√15-8]/2=√15-4

x₂=[2√15+8]/2=√15+4