okay I re-did the problem....so the square root of (x-7) + square root of x = 4.
so I squared both sides to get x-7+x=16.
simplify: 2x-7=16
add 7 to both sides 2x=23
divide each side by 2 and x=11.5
did i get it?
Hello,
$\displaystyle \sqrt{x-7} + \sqrt{x} = 4$ . Square both sides of the equation:
$\displaystyle x-7 + 2 \cdot \sqrt{x-7} \cdot \sqrt{x} + x = 16$ . Isolate the square-root on one side of the equation:
$\displaystyle 2 \cdot \sqrt{x-7} \cdot \sqrt{x} = 23 - 2x$ . Square both sides:
$\displaystyle 4(x-7) \cdot x = 529 - 92x + 4x^2$ . Expand the brackets. Collect all like terms:
$\displaystyle 4x^2 - 28x = 529 - 92x + 4x^2 \Longrightarrow 64x = 529 \Longrightarrow x = \frac{529}{64} $
Hello, iheartthemusic29!
Isolate a radical: .$\displaystyle \sqrt{x-7} \;=\;4 - \sqrt{x}$$\displaystyle \sqrt{x-7} + \sqrt{x} \;=\;4$
I squared both sides to get: .$\displaystyle x-7+x\;=\;16$ . . . . Wrong-wrong-wrong!
Square both sides . . . carefully!
. . $\displaystyle \left(\sqrt{x-7}\right)^2\;=\;(4 - \sqrt{x})^2\quad\Rightarrow\quad x - 7 \;=\;16 - 8\sqrt{x} + x \quad\Rightarrow\quad8\sqrt{x} \;= \;23$
Square both sides: .$\displaystyle \left(8\sqrt{x}\right)^2\;=\;23^2\quad\Rightarrow\ quad64x \;=\;529$
Therefore: .$\displaystyle \boxed{x \;=\;\frac{529}{64}}$