Question

1) a. 3x² - 2x-1=0
b. x² + 3x-4=0
c. x²+4x+5=0
d. x² + 4x +4=0
e. x²-2x-2=0
f. x²-3x=0
2) a. (x+3) ² = 1
b. (x-2)²=5
c. (x-3) (x+2) =0
d. (x+1)(x-2)=0
e. (x+4)(x-5) =-20
f. (x-2)(x+3)=6
3) a) x + 1/x=2
b) x²+1/x=5/2
c) 1/x+1+1/x+2=5/6
d) x-1/x+1+x-2/x+2=-1/3
4) a) x(x-3)=4
b) x(x+2)=3
c) x²+(x-1)(x+2)=1
d) (x+1)(x-2)+(x-1)(x+2)=-2
e) (x+1)²+(x-3)²=16

sunt praf la mate si acum nu imi mai amintesc nimic =)))
nu trebuie chiar toate, cat se poate

Answer 1

-la ecuatiile de gradul 2 care sunt de forma ax²+bx+c=0

se calculeaza delta=b²-4ac

daca delta>0 atunci are 2 solutii

daca delta=0 atunci are o solutie

daca delta<0 atunci nu are solutii in R

x1=(-b+radical(delta))/2a

x2=(-b-radical(delta))/2a

exemple:

1.a)3x²-2x-1=0

delta=(-2)²-4*3*(-1)

delta=4+12

delta=16

x1=(-(-2)+rad(16))/6

x1=1

x2=(2-4)/6

x2=-1/3

1.f)x²-3x=0

aici c=0

delta=(-3)²-4*1*0

delta=9

x1=(3+3)/2

x1=3

x2=(3-3)/2=0

2.b) (x-2)²=5

x²-4x+4=5

x²-4x+4-5=0

x²-4x-1=0 (aducem totul in stanga pentru a avea in dreapta 0 si a forma o ec de gradul 2)

de aici se afla delta si solutiile

binomul la patrat: (a+b)²=a²+2ab+b²

(a-b)²=a²-2ab+b²

2.f)(x-2)(x+3)=6

x*x+3x-2x-6=6

x²+x-6=6

x²+x-12=0

.....

3.a)x+1/x=2

aducem la acelasi numitor

(x²+1)/x=2

x²+1=2x

x²-2x+1=0

3.d)x-1/x+1+x-2/x+2=-1/3

(x²-1+x+x²-2+2x)/x=-1/3

3( 2x²+3x-3)=-x

6x²+9x-9=-x

6x²+8x-9=0

.....

4.d)(x+1)(x-2)+(x-1)(x+2)=-2

x²-x-2+x²+x-2+2=0

2x²-2=0 |:2

x²-1=0

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

x-1=0 =>x1=1

x+1=0 => x2=-1

a²-b²=(a-b)(a+b)