Find the positive constant k so that the equation has a unique solution in x.

SpongeBobRules24 Apr 7, 2020

#1**0 **

When plugging in values of x that are squares, you get k to equal 2.

So your answer should be **k= 2**.

Hope it helps!

HELPMEEEEEEEEEEEEE Apr 7, 2020

#3**+1 **

First square the equation and rearrange

3sqrt(x)=x+k

9x=x^{2}+2kx+k^{2}

0=x^{2}+2kx-9x+k^{2}

Factor to make it a quadratic

0=x^{2}+(2k-9)x+k^{2}

For a quadratic equation to have one solitution, the discrmimnant must be 0.

b^{2}-4ac=0

Substitute in the coefficents:

(2k-9)^{2}-4(1)(k^{2})=0

Expand

4k^{2}-36k+81-4k^{2}=0

Simplify

-36k+81=0

Solve

-4k+9=0

-4k=-9

k=2.25

power27 Apr 7, 2020