Homework Help: 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