Answer:
a-bi
Step-by-step explanation:
If a quadratic equation lx^2+mx+n=0 has one imaginary root as a+bi then the other root is the conjugate of a+bi = a-bi
Because we have l, m and n are real numbers and they are the coefficients.
Sum of roots = a+bi + second root = -m/l
When -m/l is real because the ratio of two real numbers, left side also has to be real.
Since bi is one imaginary term already there other root should have -bi in it so that the sum becomes real.
i.e. other root will be of the form c-bi for some real c.
Now product of roots = (a+bi)(c-bi) = n/l
Since right side is real, left side also must be real.
i.e.imaginary part =0
bi(a-c) =0
Or a =c
i.e. other root c-bi = a-bi
Hence proved.
Answer:
81y^2+36y+4
Step-by-step explanation:
First: 9y^2= 81y
Next: 2^2= 4
Then: 9y times 2= 18
Last: 18 times 2= 36y
(this : ^2 means squared in case you did not know)
Yes, it's an identity because remember sin 6x would be (3x + 3x)= which would make it 2 sin 3x....concluding the equation should be an identity. :)
Hoped this helped :)
Ratios equivalent to 5:1 are
10:2
15:3
