For Medal : Help please with checking if i got the correct answer with simplifying complex fractions, question below. The answer i got is -1. Thanks :)

Question

anonymous · Answer

$$\frac{ 35x ^{2} + 2x - 1 }{ 15x + 3 } \div \frac{ 2 - 98x ^{2} }{6x + 42x }$$

anonymous · Answer

Is that your answer? Or is that the question?

anonymous · Answer

that is the question, the answer i got was -1

jim_thompson5910 · Answer

is that second denominator really 6x+42x?

anonymous · Answer

sorry, it is 6 + 42x, thank you

jim_thompson5910 · Answer

the answer is not -1, try it again

anonymous · Answer

ok, thanks i will

jim_thompson5910 · Answer

oh wait, i put in 6-42x 

my bad

jim_thompson5910 · Answer

I changed it to 6+42x and it's saying -1 now, so you got it

anonymous · Answer

ok great, thanks you. do u have time to help with one more?

jim_thompson5910 · Answer

sure, go for it

anonymous · Answer

one root of a quadratic equation is -3 + 2i. Determine the other root and write an equation that would result in these roots. I'm not sure how to do this one.

jim_thompson5910 · Answer

if one root is a+bi, then the other root must be a-bi since all roots come in conjugate pairs

jim_thompson5910 · Answer

so what's the conjugate of -3+2i?

anonymous · Answer

-3 - 2i

anonymous · Answer

how do u write the equation?

jim_thompson5910 · Answer

well if the two roots are -3+2i and -3-2i, then we know that

x = -3+2i or x = -3-2i

x+3 = 2i or x+3 = -2i

(x+3)^2 = (2i)^2 or (x+3)^2 = (-2i)^2


I'll let you finish

anonymous · Answer

why are the terms in the last line squared?

jim_thompson5910 · Answer

well remember that $\Large i = \sqrt{-1}$

if you square both sides, you get $\Large i^2 = -1$

jim_thompson5910 · Answer

that's why I'm squaring both sides, to get rid of the imaginary i term

anonymous · Answer

thanks for explaining...so the answer would be x^2 + 6x + 11 or x^2 + 6x + 7?

jim_thompson5910 · Answer

neither choice, here's why:


x = -3+2i or x = -3-2i

x+3 = 2i or x+3 = -2i

(x+3)^2 = (2i)^2 or (x+3)^2 = (-2i)^2

(x+3)^2 = (2)^2*(i)^2 or (x+3)^2 = (-2)^2*(i)^2

(x+3)^2 = 4i^2 or (x+3)^2 = 4i^2

(x+3)^2 = 4i^2

(x+3)^2 = 4(-1)

(x+3)^2 = -4

x^2 + 6x + 9 = -4

x^2 + 6x + 9 + 4 = 0

x^2 + 6x + 13 = 0

anonymous · Answer

sorry, i'm still confused...in line 4 how did u get 2^2 times i^2 from 2i^2?

anonymous · Answer

wouldn't it be 2 times i^2 and then -2 times i^2

jim_thompson5910 · Answer

I used the rule that

(ab)^c = a^c * b^c 

to go from

(2i)^2

to

(2)^2 *  (i)^2

jim_thompson5910 · Answer

it's NOT 2i^2 it's really (2i)^2

jim_thompson5910 · Answer

the parenthesis make a big difference

anonymous · Answer

ok got it :) thanks so much for your help, i appreciate it!