f(x) = x^2 - 4x + 3 and g(x)= 2x - 1
solve f(x)=g(x)
x^2 - 4x + 3 = 2x - 1
-2x -2x
x^2 - 6x = -4
x^2 - x = -2/3
and now I'm stuck?
How do i get rid of x^2 - x?
Nats wrote:f(x) = x^2 - 4x + 3 and g(x)= 2x - 1
solve f(x)=g(x)
x^2 - 4x + 3 = 2x - 1
-2x -2x
x^2 - 6x = -4
x^2 - x = -2/3
Nats wrote:How do i get rid of x^2 - x?
Nats wrote:f(x) = x^2 - 4x + 3 and g(x)= 2x - 1
solve f(x)=g(x)
x^2 - 4x + 3 = 2x - 1
-2x -2x
x^2 - 6x = -4
x^2 - x = -2/3
and now I'm stuck?
How do i get rid of x^2 - x?
Nats wrote:@jg.allinsymbols, thanks again for your helpful reply.
starting from x^2 - 6x = -4.
x^2 - 6x + 4 = 0
using the quadratic formula:
-(-6) +- [(sqrt)-6^2 - 4(1)(4)]/2(1)
6+-(sqrt 20)/2
6+-(sqrt 4x5)/2
6+-2(sqrt 5)/2 and i cancel out the twos
6+-2.24
x = 8.25 or x= 3.76
is this correct?
Nats wrote:starting from x^2 - 6x = -4.
x^2 - 6x + 4 = 0
using the quadratic formula:
-(-6) +- [(sqrt)-6^2 - 4(1)(4)]/2(1)
6+-(sqrt 20)/2
6+-(sqrt 4x5)/2
6+-2(sqrt 5)/2 and i cancel out the twos
maggiemagnet wrote:But this is totally not what the Quadratic Formula gives, so I think you mean like this:
Nats wrote:from here i said the sqrt{4 is 2. so im left with : 6 plus minus 2 times the sqrt 5 and thats how i got my answers.x = 8.25 or x= 3.76