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 $x=3$ ,

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

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

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

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

$0=0$

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