Question

Find the solution to the equation: f(x)=x^2-4x+3 Please show work!

Answer 1

Let's see

$\\ \rm\rightarrowtail f(x)=0$

$\\ \rm\rightarrowtail x^2-4x+3=0$

$\\ \rm\rightarrowtail x^2-3x-x+3=0$

$\\ \rm\rightarrowtail x(x-3)-1(x-3)=0$

$\\ \rm\rightarrowtail (x-1)(x-3)=0$

$\\ \rm\rightarrowtail x=1,3$

Answer 2

Answer:

• two solutions

x = 1, 3

Explanation:

$\sf f(x)=x^2-4x+3$

$\rightarrow \sf x^2-4x+3=0$

$\rightarrow \sf x^2-3x-x+3=0$

$\rightarrow \sf x(x-3)-1(x-3)=0$

$\rightarrow \sf (x-1)(x-3)=0$

$\rightarrow \sf x-1=0, \ x-3=0$

$\rightarrow \sf x=1, \ x=3$

Thus, there are two solutions and the values are 1 and 3