Find Where Undefined/Discontinuous (k^10+9k^5+18)/(k^10+6k^5+9)

$\frac{k^{10} + 9 k^{5} + 18}{k^{10} + 6 k^{5} + 9}$

Step 1

Set the denominator in $\frac{k^{10} + 9 k^{5} + 18}{k^{10} + 6 k^{5} + 9}$ equal to $0$ to find where the expression is undefined.

$k^{10} + 6 k^{5} + 9 = 0$

Step 2

Solve for $k$ .

Factor the left side of the equation.

Rewrite $k^{10}$ as ${(k^{5})}^{2}$ .

${(k^{5})}^{2} + 6 k^{5} + 9 = 0$

Let $u = k^{5}$ . Substitute $u$ for all occurrences of $k^{5}$ .

$u^{2} + 6 u + 9 = 0$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$u^{2} + 6 u + 3^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 3 + 3^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 3$ .

${(u + 3)}^{2} = 0$

Replace all occurrences of $u$ with $k^{5}$ .

${(k^{5} + 3)}^{2} = 0$

Set the $k^{5} + 3$ equal to $0$ .

$k^{5} + 3 = 0$

Solve for $k$ .

Subtract $3$ from both sides of the equation.

$k^{5} = - 3$

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

$k = \sqrt[5]{- 3}$

Simplify $\sqrt[5]{- 3}$ .

Rewrite $- 3$ as ${(- 1)}^{5} \cdot 3$ .

Rewrite $- 3$ as $- 1 (3)$ .

$k = \sqrt[5]{- 1 (3)}$

Rewrite $- 1$ as ${(- 1)}^{5}$ .

$k = \sqrt[5]{{(- 1)}^{5} \cdot 3}$

Pull terms out from under the radical.

$k = - 1 \sqrt[5]{3}$

Rewrite $- 1 \sqrt[5]{3}$ as $- \sqrt[5]{3}$ .

$k = - \sqrt[5]{3}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$k = - \sqrt[5]{3}$

Decimal Form:

$k = - 1.24573093 \dots$

Step 4

Find Where Undefined/Discontinuous (k^10+9k^5+18)/(k^10+6k^5+9)

Related Questions

Support LCMGCF.com by disabling your adblocker.