## 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
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### 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
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Joined: Mon Dec 08, 2008 12:32 am
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### 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?

jg.allinsymbols
Posts: 72
Joined: Sat Dec 29, 2012 2:42 am
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### 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?
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
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Joined: Mon Aug 19, 2013 11:14 am
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### 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
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### Re: given f(x) = x^2 - 4x + 3 and g(x)= 2x - 1, solve f(x)=g

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

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

-(-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
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### 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
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### 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
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### 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.