Question

If A and B are roots of the equation 3x^2 + kx +9=0, then logA^2B - logA=?

Answer 1

@mathmate

Answer 2

does your question say
logA^(2B) - logA
or
log(A^2) * B - log A

Answer 3

Applying Vieta's formula to a second degree polynomial (quadratic)
P(x)=ax^2 + bx + c
the roots r1,r2 of the equation P(x)=0 satisfy

r1 + r2 = -b/a
r1*r2= c/a

Answer 4

wait, it can also be \(\large\color{black}{\log({\rm A^2B}) - \log({\rm A}) }\)

Answer 5

^

Answer 6

Applying Vieta's formula on your quadratic equation 3x^2 + kx +9 =0 we have
A + B = -k/3
A*B = 9/3 = 3
AB = 3

Now we can simplify your log expression:

log ( A^2B) - log (A)
= log(A^2) + log (B) - log A
= 2*log(A) + log (B) - log (A)
= log(A) + log(B)

but we know AB = 3

log(AB) =log(3)

and since
log(AB) = log(A) + log(B)
...
can you solve it now

Answer 7

wats the answer, I cant solve it

Answer 8

k=-9?

Answer 9

@perl

Answer 10

its not asking to solve for k

Answer 11

@perl the question says to stop at log 3 thaks

Answer 12

@CrashOnce
You have to realize that the question is lacking parentheses, and is ambiguous.
It can be one of three interpretations.
Perl has suggested two interpretations, and here's the third by @SolomonZelman .
You will not evaluate the expression until the question is clear.

You can work along the following lines if the third interpretation is valid:
or, if
\(\large\color{black}{\log({\rm A^2B}) - \log({\rm A}) }\)
as @SolomonZelman suggested, we can simplify as
\(\large=\color{black}{\log({\rm \frac{A^2B}{A}}) }\)
\(\large=\color{black}{\log({\rm AB}) }\)
\(\large=...\)
Use Vieta's formula to find the value of AB (using hints from @perl )