Question

rozwiąż równanie :

a) (x-3) (x+2) = (x+4) (2-x) - 14

b) (2-3x)do kwadratu + (3+x) (x-1) = 1+x

c) (x-3)do kwadratu + (x+5)do kwadratu = 34-x

Answer 1

a)

(x-3)(x+2) = (x+4)(2-x)-14

x²+2x-3x-6 = 2x-x²+8-4x-14

x²-x-6 = -x²-2x-6

x²+x²-x+2x-6+6 = 0

2x²+x = 0

Δ = 1²-4*2*0

Δ = 1-0

Δ = 1

√Δ = 1

x1 = (-1-1)/(2*2)

x1 = -2/4

x1 = -0,5

x2 = (-1+1)/(2*2)

x2 = 0/4

x2 = 0

 

b)

(2-3x)²+(3+x)(x-1) = 1+x

4-12x+9x²+3x-3+x²-x = 1+x

10x²-10x+1 = 1+x

10x²-11x = 0

Δ = (-11)²-4*10*0

Δ = 121-0

Δ = 121

√Δ = 11

x1 = [-(-11)-11]/(2*10)

x1 = (11-11)/20

x1 = 0/20

x1 = 0

x2 = [-(-11)+11]/(2*10)

x2 = (11+11)/20

x2 = 22/20

x2 = 1,1

 

c)

(x-3)²+(x+5)² = 34-x

x²-6x+9+x²+10x+25 = 34-x

2x²+4x+34 = 34-x

2x²+5x = 0

Δ = 5²-4*2*0

Δ = 25-0

Δ = 25

√Δ = 5

x1 = (-5-5)/(2*2)

x1 = -10/4

x1 = -2,5

x2 = (-5+5)/(2*2)

x2 = 0/4

x2 = 0