rational expressions #3 (find zeroes)

[mathymath]

i got c = 1.8 and -3 but the correct answer is c = 2 and c cannot equal +or- 3

 

[lookagain]

i got c = 1.8 and -3 but the correct answer is c = 2 and c cannot equal +or- 3

mathymath,

here is the initial step that is incorrect:


\(\displaystyle \dfrac{c + 3(c - 3)}{(c + 3)(c - 3)}\)


Certain issues:

You are multiplying each expression in the equation by the LCD, not by what (c + 3) lacks as a factor of the LCD.

Even if you were to multiply (c + 3) by (c - 3), there would be parentheses around the (c + 3) in the numerator.

Even if all of those factors were to have cancelled out, you wouldn't have 0 (nothing), but instead you would
have 1 (a type of place holder expression) where the second fraction used to be.

You are missing the chance to simplify (reduce) the second fraction at the beginning so that your LCD will be a
linear expression.


-------------------------------

Corrected work:


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{c + 3}{c^2 - 9} \ - \ 5 \ \implies\)


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{c + 3}{(c + 3)(c - 3)} \ - \ 5 \ \implies\)


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{1}{c - 3} \ - 5 \ \implies \ \ \ \ \) if c not equal to -3


\(\displaystyle \dfrac{6(c - 3)}{c - 3} \ = \ \dfrac{1(c - 3)}{c - 3} \ - \ 5(c - 3) \ \implies \ \ \ \ \) if c not equal to 3


\(\displaystyle 6 \ = \ 1 \ - \ 5c + 15\)


Can you continue from there?

 

[mathymath]

Corrected work:


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{c + 3}{c^2 - 9} \ - \ 5 \ \implies\)


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{c + 3}{(c + 3)(c - 3)} \ - \ 5 \ \implies\)


\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{1}{c - 3} \ - 5 \ \implies \ \ \ \ \) if c not equal to -3


\(\displaystyle \dfrac{6(c - 3)}{c - 3} \ = \ \dfrac{1(c - 3)}{c - 3} \ - \ 5(c - 3) \ \implies \ \ \ \ \) if c not equal to 3


\(\displaystyle 6 \ = \ 1 \ - \ 5c + 15\)

i dont understand, i know the denominator has to be (c-3)(c-3)(c+3)(1), im just not sure what should go at the top... im suppose to be left with a quadratic equation in the end

 

[lookagain]

i dont understand, i know the denominator has to be (c-3)(c-3)(c+3)(1),

im just not sure what should go at the top...

im suppose to be left with a quadratic equation in the end

If you do it my way, you won't have a quadratic equation in the end.


However, let's go back to your more complicated way:



\(\displaystyle \dfrac{6}{c - 3} \ = \ \dfrac{c + 3}{(c + 3)(c - 3)} \ - \ 5 \ \implies\)


\(\displaystyle \dfrac{6(c + 3)(c - 3)}{c - 3} \ = \ \dfrac{(c + 3)[(c + 3)(c - 3)]}{(c + 3)(c - 3)} \ - \ 5(c + 3)(c - 3) \ \implies\)


\(\displaystyle 6(c + 3) \ = \ c + 3 \ - \ 5(c + 3)(c - 3) \ \implies\)


\(\displaystyle 6(c + 3) \ = \ c + 3 \ - \ 5(c^2 - 9) \ \implies \ \ what \ ?\)

 

[mathymath]

thank you, i get it now, i see how you were doing it your way that is much faster. sorry im a little slow...