Find Where Undefined/Discontinuous (x^2-81)/(x^2)*(x^2-9x)/(x^2+x-90)

$\frac{x^{2} - 81}{x^{2}} \cdot \frac{x^{2} - 9 x}{x^{2} + x - 90}$

Set the denominator in $\frac{x^{2} - 81}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Solve for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set the denominator in $\frac{x^{2} - 9 x}{x^{2} + x - 90}$ equal to $0$ to find where the expression is undefined.

$x^{2} + x - 90 = 0$

Solve for $x$ .

Factor $x^{2} + x - 90$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 90$ and whose sum is $1$ .

$- 9, 10$

Write the factored form using these integers.

$(x - 9) (x + 10) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 9 = 0$

$x + 10 = 0$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

Set $x + 10$ equal to $0$ and solve for $x$ .

Set $x + 10$ equal to $0$ .

$x + 10 = 0$

Subtract $10$ from both sides of the equation.

$x = - 10$

The final solution is all the values that make $(x - 9) (x + 10) = 0$ true.

$x = 9, - 10$

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

$x = - 10, x = 0, x = 9$

Find Where Undefined/Discontinuous (x^2-81)/(x^2)*(x^2-9x)/(x^2+x-90)

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