Find Where Undefined/Discontinuous (32-2y^2)/(5y^2-80)

$\frac{32 - 2 y^{2}}{5 y^{2} - 80}$

Step 1

Set the denominator in $\frac{32 - 2 y^{2}}{5 y^{2} - 80}$ equal to $0$ to find where the expression is undefined.

$5 y^{2} - 80 = 0$

Step 2

Solve for $y$ .

Add $80$ to both sides of the equation.

$5 y^{2} = 80$

Divide each term in $5 y^{2} = 80$ by $5$ and simplify.

Divide each term in $5 y^{2} = 80$ by $5$ .

$\frac{5 y^{2}}{5} = \frac{80}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y^{2}}{5} = \frac{80}{5}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{80}{5}$

Simplify the right side.

Divide $80$ by $5$ .

$y^{2} = 16$

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

$y = \pm \sqrt{16}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$y = \pm \sqrt{4^{2}}$

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

$y = \pm 4$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 4$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 4$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 4, - 4$

Step 3

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 $0$ .

$y = - 4, y = 4$

Step 4

Find Where Undefined/Discontinuous (32-2y^2)/(5y^2-80)

Related Questions

Support LCMGCF.com by disabling your adblocker.