Question

Polecenie: Rozwiąż równania za pomocą równania kwantowego zupełnego.

1. (x+3)kwadrat + 6(x+3)=16
2. (3x+2)(x-1)-13=2x(2-x)
3. 9(x-4)kwadrat + 4(x-4)=0
4. (y+3)(y+2)=1-y

Answer 1

Δ=b²-4ac

x₁=[-b-√Δ]/2a

x₂=[-b-√Δ]/2a

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1. (x+3)²+ 6(x+3)=16

(x+3)[x+3+6]-16=0

(x+3)(x+9)-16=0

x²+12x+11=0

Δ=144-44=100

√Δ=10

x₁=-22/2=-11

x₂=-2/2=-1

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2. (3x+2)(x-1)-13=2x(2-x)

3x²-x-2-13-4x+2x²=0

5x²-5x-15=0

5(x²-x-3)=0

Δ=1+12=13

√Δ=√13

x₁=[1-√13]/2

x₂=[1+√13]/2

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3. 9(x-4)kwadrat + 4(x-4)=0

(x-4)[9x-36+4]=0

(x-4)(9x-32)=0

9x²-68x+128=0

Δ=4624-4608=16

√Δ=4

x₁=-72/18=-4

x₂=-64/18=-16/3

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4. (y+3)(y+2)=1-y

y²+5y+6-1+y=0

y²+6y+5=0

Δ=36-20=16

√Δ=4

y₁=-10/2=-5

y₂=-2/2=-1