How do you solve #\frac { x ^ { 2} - 8x + 15} { x - 5} = 0#?

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Answer 1 · 2017-04-24T22:55:07

#x=3#

#\frac { x ^ { 2} - 8x + 15} { x - 5} = 0#

First solve the numerator by quadratic formula or factoring method.

#[(x-5)(x-3)]/(x-5)=0#

Divide common factors

#[cancel((x-5))(x-3)]/cancel((x-5))=0#

#x-3=0#

#x=3#

We can check it too

Since #x=3#,

#\frac { x ^ { 2} - 8x + 15} { x - 5} = 0#

#\frac { 3 ^ { 2} - 8(3) + 15} { 3 - 5} = 0#

#\frac { 9 - 24+ 15} { -2} = 0#

#\frac { 0} { -2} = 0#

#0=0#