(x-3)^(1/2) - (x)^(1/2) = -3
How many solutions, if any are in this equation? I can't seem to be able to factor this :S Any suggestions are greatly appreciated!
$\displaystyle \begin{array}{rcl} \sqrt{x-3} - \sqrt{x} & = & - 3 \\ \sqrt{x - 3} & = & \sqrt{x} - 3 \\ x - 3 & = & x - 6\sqrt{x} + 9 \qquad \text{Squared both sides} \\ 6\sqrt{x} & = & 12 \\ & \vdots & \end{array}$
What can you concude?
Hello, finalfantasy!
Why are you trying to factor?$\displaystyle \sqrt{x-3} - \sqrt{x} \;=\; -3$
How many solutions, if any, are in this equation?[
We have:
. $\displaystyle \sqrt{x-3} \;=\;\sqrt{x} - 3$
Square both sides:
. . $\displaystyle x - 3 \;=\;x - 6\sqrt{x} + 9 $
. . .$\displaystyle 6\sqrt{x} \;=\;12$
. . . $\displaystyle \sqrt{x} \;=\;2$
. . . . $\displaystyle x \;=\;4\quad\hdots$ . but this root is extraneous
There are no solutions.