A polynomial equation with rational coefficients has the roots 7 + *sqrt* 3 , 2 - *sqrt* 6. Find two additional roots. 7 - *sqrt* 3 , 2 + *sqrt* 6 3 - *sqrt* 7 , 6 + *sqrt* 2 7 + *sqrt* 3 , 2 - *sqrt* 6 3 + *sqrt* 7 , 6 - *sqrt* 2
If the coefficients are rational, then those irrational roots have to come in conjugate pairs.
ummm , i have no idea what you mean by that ? lol
Ok, that's fine. Let's go over some vocabulary then. The square root of a number that is not a perfect square is irrational. 7+√3 and 2-√6 are irrational.
If the polynomial has rational coefficients, then those irrational numbers have to cancel out. That's why you need a conjugate (a number that has the opposite of what you're trying to cancel).
ok , i see . . .
e.g. the conjugate of 1+√2 is 1-√2. The +√2 and -√2 will cancel out to give rational coefficients.
yesss . .
So, of those choices, which contains the conjugates of 7+√3 and 2-√6? Look carefully, because they are all pretty similar-looking.
b ?
or c ?
Neither of those.
:( i feel slow . . . . lol
well then i have to say d now ?
Hmm, you might be misunderstanding what I mean by "opposite" "Opposite means the + or - signs are switched.
+3 is the opposite of -3, etc.
oh , i guess i did misunderstand it . it has to be a then .
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