Question

Blagam o rozwiazanie tego zdania! potrzebuje na jutro :(

e.-5x-4xdo potegi 2 =0

f.1/3x do potegi 2 =5x

g.-2/7x do pot 2 + 2x =0

h.x do pot 2 + 2/2=4x +1

i.x+9=(x-3)do pot 2

j.(4x-6)do pot 2 = (x-4)(x-9)

k.(x-2)(x+2)=5

to są równania kwadratowe ;(

Answer 1

e) -5x-4x^2=0

delta=(-5)^2-4*(-4)*0=25-0=25   pierwistek z delty=5

x1=-(-5)-5/2*(-4)=0/-8=0

x2=-(-5)+5/2*(-4)=10/-8= mines jedna cała i 2/8

 

f)1/3x^2=5x

1/3x^2-5x=0

delta=(-5)^2-4*1/3*0=25-0=25    pierwsiatek to 5

x1=-(-5)-5/2*1/3=0/2/3=0

x2=-(-5)+5/2/3=10/2/3=6 całych i 2/3

 

g)-2/7x^2+2x=0

delta=2^2-4*(-2/7)*0=4-0=4

x1=-2-2/2*(-2/7)=-4/-4/7=-7

x2=-2+2/2(-2/7)=0

 

h)x^2+2/2=4x+1

x^2+2/2-4x-1=0

x^2-4x=0

delta= (-4)^2-4*1*0=14-0=16 pierwiastek =4

x1=-(-4)-4/2*1=0

x2=-4+4/2*1=0

 

i)x+9=(x-3)^2

x+9=[x^2+2*x*3-3^2]

x+9=x^2+6x-9

x+9-x^2-6x+9=0

-x^2-5x+18=0

delta=(-5)^2-4*(-1)*18=25+72=97

x1=-(-5)-97/2*(-1)=(-92)/(-2)=46

x2=-(-5)+97/2*(-1)=102/(-2)=-51

 

k)(X-2)(x+2)=5

x^2-2^2=5

x^2-4=5

x^2-9=0

delta=0-4*1*-9=36   pierwiastek =6

x1=-6/2*1=-3

x2=6/2*1=3