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:
|x-3|-5
Step-by-step explanation:
f(x) = |x|
These lines are lines of absolute value equations. g(x) is shifted 3 units left so we subtract 3 from within the bars, and shifted down 5 units so we subtract 5 from the absolute value which leaves us with |x-3|-5
Answer:
the answer is D
Step-by-step explanation:
i got it right
Yo sup??
since WY is the perpendicular bisector we can say
3x-5=2x+3
x=8
Hope this helps
