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

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

Postby Nats » Thu Aug 22, 2013 1:20 pm

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?

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

Postby maggiemagnet » Thu Aug 22, 2013 6:38 pm

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?

You don't. You solve the quadratic! Probably with the Quadratic Formula.
:clap:

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

Postby jg.allinsymbols » Thu Aug 22, 2013 7:51 pm

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?


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

Postby Nats » Fri Aug 23, 2013 11:09 am

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

Postby Nats » Fri Aug 23, 2013 11:11 am

@maggie, thank you for your reply as well.

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

Postby jg.allinsymbols » Sun Aug 25, 2013 4:15 am

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?


Your grouping symbols are misplaced.

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

Postby maggiemagnet » Sun Aug 25, 2013 12:14 pm

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

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



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









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

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

Postby Nats » Mon Aug 26, 2013 10:45 am

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








yes. 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

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

Postby buddy » Mon Aug 26, 2013 3:59 pm

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

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

Postby Nats » Mon Aug 26, 2013 4:20 pm

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


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