# Find the value of k

1. Dec 10, 2011

### darshanpatel

1. The problem statement, all variables and given/known data

For the equation 4(times)sqrtx= 2x+k find a value such that k the equation has (a) one solution, (b) two solutions, (c) no solutions

2. Relevant equations

None

3. The attempt at a solution

From original equation, I got to:

sqrtx=(2x+k)/4

x=((2x+k)/4)^2

x= (4x^2+4xk+k^2)/16

simplified to: x= 1/4x^2 +1/4xk + 1/16k^2

Put it into quadratic formula and got the discriminant as: sqrt((1/4k)^2 -(4)(1/4)(k^2/16))

2. Dec 10, 2011

### SammyS

Staff Emeritus
You don't have the equation, x= 1/4x^2 +1/4xk + 1/16k^2, in the correct form to use the quadratic formula. There's an x on the left hand side.

More simply:
Square the original equation, $\displaystyle 4\sqrt{x}= 2x+k$

giving $\displaystyle 16x= 4x^2+4kx+k^2$

$\displaystyle 4x^2+(4k-16)x+k^2=0$​

Use the discriminant, b2-4ac, to determine the value of k needed for 0, 1, or 2 real solutions.

3. Dec 10, 2011

### darshanpatel

Oh, thanks, So for 1 solution the discriminant would have to be 0, and for 2 solutions ≤0 and what would it have to be for 0 solutions? just ≥0? becuase cant u still get 2 imaginary solutions?

4. Dec 10, 2011

### darshanpatel

the final answer I got was: k=25/128 for one solution k>25/128 for no solutions and k<25/128 for 2 real solutions

5. Dec 10, 2011

### eumyang

No, not quite.
If b2-4ac = 0, there is 1 real solution.
If b2-4ac > 0, there are 2 real solutions.
If b2-4ac < 0, there are no real solutions (but there are 2 complex solutions).

6. Dec 10, 2011

### darshanpatel

oh yeah, sorry, I accidentally included the "equal to" but I fixed it in the final answer..

7. Dec 10, 2011

### eumyang

The "equal to" is not the only thing that's incorrect in the bolded part above.

8. Dec 10, 2011

### darshanpatel

oh, it wasnt bolded in previous part, but i see, >0 is 2 solutions, <0 no solutions( complex) = 0 one solution