real values

[maeveoneill]

If f(x) = 2x -1, determine all real values of x such that (f(x))^2 -3f(x) +2 =0.

This is what I have gotten:

(2x-1)^2 - 3(2x-1) +2 = 0
4x^2 -4x +1 -6x +3 +2 =0
4x^2 -10x +3 =0

Using the quadratic formula to find x, I got:
x = 10 +/- sqrt52
. --------------
. 8

Therefore, the real value(s) of x are:
x = 5 +/- sqrt13
. --------------
. 2

Can someone tell me if this is right?

 

[soroban]

Hello, maeveoneill!

If $\displaystyle f(x)\:=\:2x\,-\,1$, determine all real values of $\displaystyle x$
such that: $\displaystyle \,[f(x)]^2\,-\,3[f(x)]\,+\,2\;=\;0$

I think they expected us to notice that we can factor immediately:

$\displaystyle \;\;[f(x)\,-\,1]\,[f(x)\,-\,2]\;=\;0$

(1) Then: $\displaystyle \,f(x)\,-\,1\:=\:0\;\;\Rightarrow\;\;f(x)\,=\,1\;\;\Rightarrow\;\;2x\,-\,1\:=\:1\;\;\Rightarrow\;\;x\,=\,1$

(2) And: $\displaystyle \,f(x)\,-\,2\:=\:0\;\;\Rightarrow\;\;f(x)\,=\,2\;\;\Rightarrow\;\;2x\,-\,1\:=\:2\;\;\Rightarrow\;\;x\,=\,\frac{3}{2}$

 

[maeveoneill]

yep youre answer im sure is right. I just dont understand something thoughhh, was the way I did it wrong? Although its a longer way, shouldnt it have come up with the same answeR?

 

[Gene]

(2x-1)^2 - 3(2x-1) +2 = 0 OK
4x^2 -4x +1 -6x +3 +2 =0 OK
4x^2 -10x +3 =0 1+3+2=6
4x^2 -10x +6 =0
Try the quadratic formula on that

Using the quadratic formula to find x, I got:
x = 10 +/- sqrt52
. --------------OK but
. 8

Therefore, the real value(s) of x are:
x = 5 +/- sqrt13
. -------------- 10/8 is not = 5/2
. 2