hi just wondering can anybody solve 3=x^2-5+2x by factorising? ?

x² + 2x - 5 = 3
x² + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = -4, 2

Answer: x = -4, 2

x² + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = - 4 , x = 2

3=(x+1)^2-6
9=(x+1)^2
(x+1)^2-9=0
(x+1-3)(x+1+3)=0
(x-2)(x+4)=0
x-2=0 s0 x=2
x+4=0 so x=-4

x^2+2x_8=0
(x-2)(x+4)=0
x=2, -4

x^2 + 2x -5 = 3
Therefore x^2 + 2x - 8 = 0 (bringing the three over)
Factorizing that:

(x + 4) (x - 2) = 0
x = -4 or x = 2

3 = x^2 - 5 - 2x
x^2 - 2x - 5 - 3 = 0
x^2 - 2x - 8 = 0
x^2 + 2x - 4x - 8 = 0
(x^2 + 2x) - (4x + 8) = 0
x(x + 2) - 4(x + 2) = 0
(x + 2)(x - 4) = 0

x + 2 = 0
x = -2

x - 4 = 0
x = 4

∴ x = -2 , 4

EL CRITERIO BÁSICO SE LLAMA COMPOSICIÓN DE
FUNCIONES

f(x)= ax+b

g(x)= ax+b entonces: f(x)=g(x)

( f o g ) = g ( f(x) )= { g( ax+b ) } = a (ax+b) + b

(f o g )= (a^2)x + ab +b

por lo tanto : ( f o g ) = ( g o g ) = g ( g(x) ) = (a^2)x + ab +b

g ( g(x) ) = (a^2)x + ab +b ;

----------
g ( g(x) ) = 6x-8 ; (a^2)x=6x ; a^2 = ( 6x) / x ; a^2=6

( a^2 )^ (1/2) = ( 6 )^(1/2) ; a= ( 6 )^(1/2)

b= - 8

entonces : g(x)= ax+b= [ a ]x + b

g(x)= [ ( 6 )^(1/2) ]x +( - 8 )

g(x)= [ ( 6 )^(1/2) ]x - 8

juan fernando gutiérrez mejía Díaz Gonzalez
pereira , risaralda.
cc 79415265
fundación universitaria del area andina , sede pereira , risaralda, colombia

-----------------------

Question
hi just wondering can anybody solve 3=x^2-5+2x by factorising? ?
The method is the part that is important to me! Thanks

organizamos primero :

x^2+2x-5=3 ; x^2+2x-5-3=0 ; x^2+2x-8=0

x1=(-b+p) / 2a ; x2=(-b-p) / 2a

p= [ (b)^2 - 4ac ] ^ (1/2)

p= [ - 28 ] ^(1/2) = { [28]^(1/2)} * i cuando i= (- 1)^(1/2)

... x1=( -2+ [ { [28]^(1/2)} * i] ) / 2

.... x1= -1 + [ { [28]^(1/2)} * i] / 2

..... x2= -1- [ { [28]^(1/2)} * i] / 2

x1=x sub 1 ....; x2=x sub 2

las respuestas pertenecen a los números complejos
-----------

3=x^2-5+2x
x^2 +2x - 8 =0
(x + 4)(x - 2) =0
x= 2 or - 4

No sorry.