Question

If 1/4-3i is a root of ax²+bx+1=0,where a,b are real,then find the values of a and b

solution ::-
1/4-3i= 4+3i/4+3i
1/4-3i is 4-3i/25
so, 4-3i/25is another root
sum ::- 1/4-3i +4-3i/25=8/25
so, the equation is the
x²-8/25x +1/25=0
or, 25x²-8x+1=0
we get = A=25,B=-8​

Answer 1

Answer:

1/4-3i= 4+3i/4+3i

1/4-3i is 4-3i/25

so, 4-3i/25is another root

sum ::- 1/4-3i +4-3i/25=8/25

so, the equation is the

x²-8/25x +1/25=0

or, 25x²-8x+1=0

we get = A=25,B=-80607-{94906