# Math Help - Radical Equation

I could not make any headway on this problem. Thanks for viewing and helping.

Using:
4 √x = 2x + k
find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.

2. CORRECTED VERSION

1. Square both sides of the equation to get:

2. Set the equation = 0

3. Factor x from the two middle terms:

4. Find the discriminant. From the quadratic formula, the discriminant is what's under the radical, namely:

You can determine the nature of the roots by inspecting the discriminant (D). If D=0, then you have one real root; if D>0, then you have 2 real roots, and if D<0, you have no real roots.

I found the discriminant to be:

5. Simplifying:

If D=0, then k=2, meaning you have one real root.

If D>0, then 0<k<2, meaning you have two real roots.

3. Originally Posted by VAP
I could not make any headway on this problem. Thanks for viewing and helping.

Using:
4 √x = 2x + k
find three different expressions to substitute for k so that the equation has two, one, and no solutions. Describe how you found the equations.
$4 \sqrt{x}=2x+k$

Note that $x=(\sqrt{x})^2$

So you can substitute $y=\sqrt{x}$.

The equation is now $4y=2y^2+k$

Divide by 2, to try to get a known form of a quadratic :

$y^2-2y+\frac k2=0$

Completing the square, it will give :

$y^2-2y+1+\left(\frac k2-1\right)=0$

$(y-1)^2=1-\frac k2$

It has only one solution if (y-1)^2=0... So $k=\dots$

It has no solution if the right hand side is negative (because it's not possible that a square is negative). So $k=\dots$, for example

It's ok if the right hand side is positive, but be careful, because y has to be positive (since $\sqrt{x}>0$). If I thought correctly, taking k>0 and such that 1-k/2>0, it will be ok

I tried to make it without "discriminants" because I don't know if you have learnt this So it's not necessarily accurate, especially with the last part, for finding 2 solutions.