## given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g(x)

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g(x)

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

Posts: 18
Joined: Mon Aug 19, 2013 11:14 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

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

How did you get your last line here? What happened to the 6 and the 4?

Nats wrote:How do i get rid of x^2 - x?

maggiemagnet

Posts: 302
Joined: Mon Dec 08, 2008 12:32 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

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?

Try to remove all terms from one side and put them onto the other side, using the inverses of the terms. Simplify the resulting expression.
..., in fact, I see you mistake. You improperly divided almost both sides by 3. That is no help. Go back to the line, x^2-6x=-4 and continue from this. You would want to get all terms on one side of the equation and you will have a quadratic equation. From there, use general solution to a quadratic formula.

You should not ask, "how do I get rid of x^2-x". You should not want to get rid of it. That is part of the quadratic expression in the resulting quadratic equation which you would solve through the general solution.
jg.allinsymbols

Posts: 71
Joined: Sat Dec 29, 2012 2:42 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

starting from x^2 - 6x = -4.
x^2 - 6x + 4 = 0

-(-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

Posts: 18
Joined: Mon Aug 19, 2013 11:14 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

Nats

Posts: 18
Joined: Mon Aug 19, 2013 11:14 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

starting from x^2 - 6x = -4.
x^2 - 6x + 4 = 0

-(-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?

jg.allinsymbols

Posts: 71
Joined: Sat Dec 29, 2012 2:42 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

Nats wrote:starting from x^2 - 6x = -4.
x^2 - 6x + 4 = 0

-(-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

The way you've written it, your first line means like this:

$-(-6)\, \pm\, \frac{\sqrt{()}\, -\, 6^2\, -\, 4(1)(4)}{2(1)}$

But this is totally not what the Quadratic Formula gives, so I think you mean like this:

$\frac{-(-6)\, \pm\, \sqrt{(-6)^2\, -\, 4(1)(4)}}{2(1)}$

$\frac{6\, \pm\, \sqrt{36\, -\, 16}}{2}$

$\frac{6\, \pm\, \sqrt{20}}{2}$

$\frac{6\, \pm\, \sqrt{4\times 5}}{2}$

But how did you "cancel out the twos" and still have a 6 on top?

maggiemagnet

Posts: 302
Joined: Mon Dec 08, 2008 12:32 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

maggiemagnet wrote:But this is totally not what the Quadratic Formula gives, so I think you mean like this:

$\frac{-(-6)\, \pm\, \sqrt{(-6)^2\, -\, 4(1)(4)}}{2(1)}$

$\frac{6\, \pm\, \sqrt{36\, -\, 16}}{2}$

$\frac{6\, \pm\, \sqrt{20}}{2}$

$\frac{6\, \pm\, \sqrt{4\times 5}}{2}$

yes. from here $\frac{6\, \pm\, \sqrt{4\times 5}}{2}$ 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
Nats

Posts: 18
Joined: Mon Aug 19, 2013 11:14 am

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

Nats wrote:from here $\frac{6\, \pm\, \sqrt{4\times 5}}{2}$ 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

what abt the 2 on the bottom?
buddy

Posts: 141
Joined: Sun Feb 22, 2009 10:05 pm

### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

I cancelled it out with the 2 I got from the sqrt of 4.
Nats

Posts: 18
Joined: Mon Aug 19, 2013 11:14 am