Question

rozwiąż nierownosci kwadratowa

x(x+2)>4(X-2)

x(x-2)>2(x-2)

(3x-2)*=(2x+3)*

(X-2)(x+2)=3x

x(x+5)=10(x+5)

(2x+1)(X-2)=0

x*+x-6+0

(8x+4)(10x+2)=0

(x+5(x-10)=0

Answer 1

1)x(x+2)>4(X-2)

x² +2 > 4x -8

x² - 4x + 10 > 0

Δ= 16 - 40 = -36

x∈R

2)x(x-2)>2(x-2)

x²-2x > 2x -4

x²-4x - 4>0

Δ=16+16=32

√Δ=4√2

x1=4-4√2/2=2-2√2

x2=4+4√2/2=2+2√2

x∈(-∞,2-2√2)∨(2+2√2,+∞)

3)(3x-2)*=(2x+3)*

9x²-12x+4=4x²+12x+9

5x²-24x-5=0

Δ=576+100=676

√Δ=26

x1=24-26/10=-1/5

x2=24+26/10=5

 

4)(X-2)(x+2)=3x

x²-3x-4=0

Δ=9+16=25

√Δ=5

x1=3-5/2=-1

x2=3+5/2=4

 

5)x(x+5)=10(x+5)

x²+5x=10x+50

x²-5x-50=0

Δ=25+200=225

√Δ=15

x1=5-15/2=-5

x2=5+15/2=10

 

6)(2x+1)(X-2)=0

2x²-4x+x-2=0

2x²-3x-2=0

Δ=9+16=25

√Δ=5

x1=3-5/4=-1/2

x2=3+5/4=2

7)x*+x-6=0

Δ=1+24

√Δ=5

x1=-1-5/2-3

x2=-1+5/2=2

8)(8x+4)(10x+2)=0

80x²+16x + 40x+8=0

80x²+56x+8=0

Δ=3136 - 2560 = 576

√Δ=24

x1=-56-24/160=-1/2

x2=-56+24/160=-32/160=-1/5

9)(x+5)(x-10)=0

x²-10x+5x-50=0

x²-5x-50=0

Δ=25+200=225

√Δ=15

x1=5-15/2=-5

x2=5+15/2=10