Solve frac{sqrt{-36}}{sqrt{-2}r-9}

$$\frac { \sqrt { - 36 } } { \sqrt { - 2 } r - 9 }$$

Evaluate (complex solution)

$$\frac{6i}{\sqrt{2}ir-9}$$

Short Solution Steps

Calculate the square root of -36 and get 6i.

$$\frac{6i}{\sqrt{-2}r-9}$$

Factor -2=2\left(-1\right). Rewrite the square root of the product \sqrt{2\left(-1\right)} as the product of square roots \sqrt{2}\sqrt{-1}. By definition, the square root of -1 is i.

$$\frac{6i}{\sqrt{2}ir-9}$$

Rationalize the denominator of \frac{6i}{\sqrt{2}ir-9} by multiplying numerator and denominator by \sqrt{2}ir+9.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{\left(\sqrt{2}ir-9\right)\left(\sqrt{2}ir+9\right)}$$

Consider \left(\sqrt{2}ir-9\right)\left(\sqrt{2}ir+9\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{\left(\sqrt{2}ir\right)^{2}-9^{2}}$$

Expand \left(\sqrt{2}ir\right)^{2}.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{\left(\sqrt{2}\right)^{2}i^{2}r^{2}-9^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{2i^{2}r^{2}-9^{2}}$$

Calculate i to the power of 2 and get -1.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{2\left(-1\right)r^{2}-9^{2}}$$

Multiply 2 and -1 to get -2.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{-2r^{2}-9^{2}}$$

Calculate 9 to the power of 2 and get 81.

$$\frac{6i\left(\sqrt{2}ir+9\right)}{-2r^{2}-81}$$

Use the distributive property to multiply 6i by \sqrt{2}ir+9.

$$\frac{6i\sqrt{2}ir+54i}{-2r^{2}-81}$$

Multiply 6i and i to get -6.

$$\frac{-6\sqrt{2}r+54i}{-2r^{2}-81}$$

Evaluate

$$\text{Indeterminate}$$

Short Solution Steps

Rationalize the denominator of \frac{\sqrt{-36}}{\sqrt{-2}r-9} by multiplying numerator and denominator by \sqrt{-2}r+9.

$$\frac{\sqrt{-36}\left(\sqrt{-2}r+9\right)}{\left(\sqrt{-2}r-9\right)\left(\sqrt{-2}r+9\right)}$$

Consider \left(\sqrt{-2}r-9\right)\left(\sqrt{-2}r+9\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{-36}\left(\sqrt{-2}r+9\right)}{\left(\sqrt{-2}r\right)^{2}-9^{2}}$$

Expand \left(\sqrt{-2}r\right)^{2}.

$$\frac{\sqrt{-36}\left(\sqrt{-2}r+9\right)}{\left(\sqrt{-2}\right)^{2}r^{2}-9^{2}}$$

Calculate \sqrt{-2} to the power of 2 and get -2.

$$\frac{\sqrt{-36}\left(\sqrt{-2}r+9\right)}{-2r^{2}-9^{2}}$$

Calculate 9 to the power of 2 and get 81.

$$\frac{\sqrt{-36}\left(\sqrt{-2}r+9\right)}{-2r^{2}-81}$$

Use the distributive property to multiply \sqrt{-36} by \sqrt{-2}r+9.

$$\frac{\sqrt{-36}\sqrt{-2}r+9\sqrt{-36}}{-2r^{2}-81}$$