Question

the quadratic 8 sec^2x-6secx+1=0 has ............ roots .....i need stepwise answer

Answer 1

replace sec with a variable like a and solve like any other quadratic

Answer 2

first solve this for u
8u^2-6u+1=0

Answer 3

i got 1/4 and 1/2 now what???

Answer 4

ok we replaced secx with u
since u=1/4 or u=1/2
then
secx=1/4 or secx=1/2

Answer 5

solve these for x and you are done

Answer 6

so what is final answer??
how many roots r thr???

Answer 7

if it helps you can write in terms of cosine
\[secx=\frac{1}{cosx}=> \frac{1}{cosx}=\frac{1}{4}, \frac{1}{cosx}=\frac{1}{2}=> cosx=4, cosx=2\]

Answer 8

but it cant be possible

Answer 9

right

Answer 10

becaise cosx is btw -1 and 1

Answer 11

0 roots???

Answer 12

-1<cos x<1

Answer 13

so we have no solution for quadratic equation

Answer 14

:)

Answer 15

no roots u mean???

Answer 16

of course

Answer 17

yes

Answer 18

wow thx
i giv medal

Answer 19

to both of u

Answer 20

thanks

Answer 21

:)

Answer 22

:) more qns coming up..............

Answer 23

yes:)

Answer 24

:)