Question

1/(2k-1) = (k+2)/25=4/(6k+2)
Find K.!!!
I could solve till this step, please help me proceed

Answer 1

if there really 2 equal signs?

Answer 2

Ya!! They have to be equated to find k! I am trying cross multiplication method but am not getting the answer!

Answer 3

1/(2k-1) = (k+2)/25
(2k-1)(k+2)=25
2k^2+3k-27=0

note if the third equation is equal to the other two, then all we need to solve for is this^^^

Answer 4

2(k^2+(3/2)k)= 27
2(k+(3/4))^2=27+18/16
(k+(3/4))^2= 27/2+18/32
k = +-sqrt(27/2+9/16)-(3/4)

Answer 5

LHS evaluated at K
1/(2(sqrt(27/2+9/16)-(3/4))-1) = 1/5
RHS evaluated at K
sqrt(27/2+9/16)-(3/4) = 1/5
I let K be the positive roots, you can check the negative, and since solving the 2nd,3rd equation is a quadratic, we know it will have no more than two solutions. so they are equal

Answer 6

so all three are equal for k = +-sqrt(27/2+9/16)-(3/4)

Answer 7

ThanQ!

Answer 8

do you understand?

Answer 9

I just completed the square to solve the equation, if you are not good at that, use what ever method you want.

Answer 10

Yeah!! Now il take square root and solve for k! THanks!

Answer 11

@zzr0ck3r - I am getting \[\sqrt{213/16}\] correct?

Answer 12

@zzr0ck3r - Is that correct? I am not getting a round number!

Answer 13

there are two numbers 3 and I forget the other, use a calculator, they are up there^^^

Answer 14

Okay!

Answer 15

sqrt(27/2+9/16)-(3/4)=3
-sqrt(27/2+9/16)-(3/4)=-4.5 sorry was on the phone.