## anonymous one year ago Rationalize the denominator of square root of negative 9 over open parentheses 4 minus 7 i close parentheses minus open parentheses 6 minus 6 i close parentheses. quantity of negative 3 minus 6 i over 5 quantity of 39 minus 6 i over negative 9 quantity of negative 39 minus 6 i over 17 quantity of 3 minus 6 i over 3

1. mathstudent55

Please write the problem using numbers and mathematical symbols instead of words. Use the equation editor, a copy or a scan of the original problem, or the draw tool.

2. anonymous

$\sqrt{-9}\div(4-7i)-(6-6i)$

3. anonymous

@Nnesha @mathstudent55 @nincompoop @Jhannybean

4. anonymous

@DanJS

5. misty1212

HI!!

6. mathstudent55

$$\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i)$$

7. misty1212

$\frac{\sqrt{-9}}{4+7i}$??

8. mathstudent55

To rationalize a denominator, multiply by a fraction which is the conjugate over the conjugate. Also, let's take care of the square root of the negative number.

9. mathstudent55

$$\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i)$$ $$=\dfrac{3i}{4 - 7i} \times \dfrac{4 + 7i}{4 + 7i} - (6 - 6i)$$ Do you follow so far?

10. anonymous

yes i am flowing

11. mathstudent55

Now we actually do the multiplication.

12. mathstudent55

$$=\dfrac{3i(4 + 7i)}{(4 - 7i)(4 + 7i)} - (6 - 6i)$$ $$=\dfrac{12i + 21i^2}{16 - (7i)^2} - (6 - 6i)$$ $$=\dfrac{12i - 21}{16 + 49} - (6 - 6i)$$

13. mathstudent55

$$=\dfrac{12i - 21}{65} - (6 - 6i)$$

14. mathstudent55

$$=\dfrac{12i - 21}{65} - \dfrac{65}{65}(6 - 6i)$$ $$=\dfrac{12i - 21}{65} - \dfrac{390-390i}{65}$$ $$=\dfrac{12i - 21- (390-390i)}{65}$$ $$=\dfrac{12i - 21- 390+390i}{65}$$ $$=\dfrac{-411 + 402i}{65}$$ $$=-\dfrac{411}{65} + \dfrac{402}{65}i$$