What are the excluded values for $(5x+1)/(x^2-1)$ ?

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Answer 1 · 2017-04-04T12:23:39

See the entire solution process below:

Because we cannot divide by $0$ the excluded values are:

$x^2 - 1 != 0$

We can factor $x^2 - 1$ using the rule:

$a^2 - b^2 = (a + b)(a - b)$

Letting $a^2 = x^2$ , $a =x$ , $b^2 = 1$ and $b = 1$ and substituting gives:

$(x + 1)(x - 1) != 0$

Now, solve each term for $0$ to find the excluded values of $x$ :

Solution 1)

$x + 1 = 0$

$x + 1 - color(red)(1) = 0 - color(red)(1)$

$x + 0 = -1$

$x = -1$

Solution 2)

$x - 1 = 0$

$x - 1 + color(red)(1) = 0 + color(red)(1)$

$x - 0 = 1$

$x = 1$

The excluded values are: $x = -1$ and $x = 1$

What are the excluded values for (5x+1)/(x^2-1)(5x+1)/(x^2-1)?