## 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.
Nats
Posts: 18
Joined: Mon Aug 19, 2013 11:14 am
Contact:

### 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?

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
Contact:

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

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?
How do i get rid of x^2 - x?
You don't. You solve the quadratic! Probably with the Quadratic Formula.

jg.allinsymbols
Posts: 72
Joined: Sat Dec 29, 2012 2:42 am
Contact:

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

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?
Your steps are wrong.
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.

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

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

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

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

@maggie, thank you for your reply as well.

jg.allinsymbols
Posts: 72
Joined: Sat Dec 29, 2012 2:42 am
Contact:

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

@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?
Your grouping symbols are misplaced.

maggiemagnet
Posts: 358
Joined: Mon Dec 08, 2008 12:32 am
Contact:

### 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

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
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?

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

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

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

buddy
Posts: 197
Joined: Sun Feb 22, 2009 10:05 pm
Contact:

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

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?

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

### 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.

Return to “Intermediate Algebra”