How do you simplify #\frac { x ^ { 2} - 6x + 8} { x - 1} = 0#?

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Answer 1 · 2018-05-04T11:45:20

#((x-4)(x-2))/(x-1)=0#

Starting out with the equation,

#((x-4)(x-2))/(x-1)=0#

Multiplying everything out

#(x^2-6x+8)/(x-1)=0#

You can see that the counter in the fraction can be factorized. So we can focus on,

#x^2-6x+8#

And try to factorize this.

There is several ways to go with this. Usually, the first one learn is the quadratic equation to help us solve this. So we can use that.

The quadratic equation looks like,

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Now we only need to figure out what #a=#, #b=# and #c=#. To do this we can read the original equation we are focusing on as,

#ax^2+bx+c#
#(x^2)+(-6x)+(8)#

From that we can see that #a=1#, #b=-6# and #c=8#. Now we can plot in the numbers into the quadratic equation,

#x=(-(-6)+-sqrt((-6)^2-4*1*8))/(2*1)#

This will give us,

#x=(6+-sqrt(36-32))/(2)=(6+-sqrt(4))/(2)=(6+-2)/(2)#

Now we have to do calculations for both,

#x_1=(6+2)/(2)#

And,

#x_2=(6-2)/(2)#

Which will be,

#x_1=(6+2)/(2)=(8)/(2)=4#

And,

#x_2=(6-2)/(2)=(4)/(2)=2#

So the #x# values will be equal to,

#x=4, x=2#

We now have the focused part factorized by writing it as,

#(x-4)(x-2)#

So we can put this into the original equation,

#((x-4)(x-2))/(x-1)=0#