Question

rozwiąż równanie kwdratowe.

1.(2x-3)^-3(x^-3x+4)=7

2.2(x+4)^-5(3x+7)=x^+3

3.3x^-5x+48=2(x+2)^

4.(x+3)^+5x^=(x-2)^

5.(x-6)^+x^=x^-17x+30

6.(2x)^-(x-5)^=0

7.25-(2x+3)^=0

Answer 1

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;

a²-b²=(a-b)(a+b) - różnica kwadratów;

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1.(2x-3)²-3(x²-3x+4)=7

4x²-12x+9-3x²+9x-12-7=0

x²-3x-10=0

Δ=9+40=49

√Δ=7

x₁=[3-7]/2=-2

x₂=[3+7]/2=5

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2.2(x+4)²-5(3x+7)=x²+3

2x²+16x+32-15x-35-x²-3=0

x²+x-6=0

Δ=1+24=25

√Δ=5

x₁=[-1-5]/2=-3

x₂=[-1+5]/2=2

====================

3.3x²-5x+48=2(x+2)²

3x²-5x+48-2x²-8x-8=0

x²-13x+40=0

Δ=169-160=9

√Δ=3

x₁=[13-3]/2=5

x₂=[13+3]/2=8

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4.(x+3)²+5x²=(x-2)²

x²+6x+9+5x²-x²+4x-4=0

5x²+10x+5=0

5(x²+2x+1)=0

5(x+1)²=0

x+1=0

x=-1 - pierwiastek podwójny

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5.(x-6)²+x²=x²-17x+30

x²-12x+36+x²-x²+17x-30=0

x²+5x+6=0

Δ=25-24=1

√Δ=1

x₁=[-5-1]/2=-3

x₂=[-5+1]/2=-2

====================

6.(2x)²-(x-5)²=0

[2x-(x-5)][2x+(x-5)]=0

[x+5][3x-5]=0

x=-5  lub  x=5/3

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7.25-(2x+3)²=0

[5-(2x+3)][5+(2x+3)]=0

[2-2x][8+2x]=0

x=1  lub  x=-4