2. Originally Posted by quebec567
You need to go through these adding brackets to make the meaning clear, or
use equation editor to make the root signs extend over what you want the
root of.

1. $\root{4}\of{x-3}=2$

raise both sides of this to the fourth power:

${x-3}=2^4=16$,

so:

$x=16+3=19$

RonL

3. Originally Posted by quebec567
Hello,

2) the same method as CaptainBlack has demonstrated:

$\sqrt[4]{y+3}=3$ will become:

$y+3=81 \Longleftrightarrow y = 78$

5) $5=\frac{1}{\sqrt{x}} \Longleftrightarrow \sqrt{x}=\frac{1}{5} \Longrightarrow x=\frac{1}{25}$

6) $\frac{2}{\sqrt{x}}=5 \Longleftrightarrow \frac{2}{5}=\sqrt{x} \Longrightarrow x=\frac{4}{25}$

7) $\sqrt{3x+1}=\sqrt{2x+6} \Longrightarrow 3x+1=2x+6 \Longrightarrow x = 5$.
You have to check this answer by re-substituting the result into the original equation because squaring an equation could change something false into a true equation.

8) and 9) are similar to 7). I leave them for you.

10) $\sqrt{x^2+6}+x-3=0$ Isolate the square-root at the LHS of the equation:

$\sqrt{x^2+6}=-x+3 \Longrightarrow x^2+6=x^2-6x+9 \Longrightarrow -3=-6x \Longrightarrow x=\frac{1}{2}$

EB

4. Hello, quebec567!

Captain Black is absolutely right.
. . Unless you use parentheses, we have to guess what you meant.

Here are a few more . . .

#2 is the same as #1
$3)\;\;\sqrt[3]{6x + 9} + 9 \:=\:6$
Subtract 9 from both sides: . $\sqrt[3]{6x + 9}\:=\:-3$

Cube both sides: . $\left(\sqrt[3]{6x + 9}\right)^3\;=\;(-3)^3$

We have: . $6x + 9 \:=\:-27\quad\Rightarrow\quad 6x\:=\:-36\quad\Rightarrow\quad x \,=\,-6$

#4 is the same as #3.
$5)\;\;5 \:=\:\frac{1}{\sqrt{x}}$
We have: . $\sqrt{x}\:=\:\frac{1}{5}$

Square both sides: . $\left(\sqrt{x}\right)^2\:=\:\left(\frac{1}{5}\righ t)^2\quad\Rightarrow\quad x\,=\,\frac{1}{25}$

#6 is the same as #5.
$8)\;\;\sqrt{5x+3} - \sqrt{2x+3}\:=\:0$
We have: . $\sqrt{5x+3} \:=\:\sqrt{2x+3}$

Square both sides: . $\left(\sqrt{5x+3}\right)^2\:=\:\left(\sqrt{2x+3}\r ight)^2\quad\Rightarrow\quad5x + 3\:=\:2x+3$

Therefore: . $3x\,=\,0\quad\Rightarrow\quad x\,=\,0$

$12)\;\;\sqrt{x-6}\:=\:\sqrt{3} - \sqrt{x}$
Square both sides: . $\left(\sqrt{x-6}\right)^2\:=\:\left(\sqrt{3} - \sqrt{x}\right)^2$

and we have: . $x - 6 \:=\:3 - 2\sqrt{3x} + x\quad\Rightarrow\quad 2\sqrt{3x}\:=\:9\quad\Rightarrow\quad \sqrt{3x}\:=\:\frac{9}{2}$

Square both sides: . $\left(\sqrt{3x}\right)^2\:=\:\left(\frac{9}{2}\rig ht)^2\quad\Rightarrow\quad3x\:=\:\frac{81}{4}$

Divide by 3: . $x\:=\:\frac{27}{4}$

5. What if the equation contains a principal root? like Number 9:
9.) 3√2x+3 = √y+10

6. Originally Posted by quebec567
What if the equation contains a principal root? like Number 9:
9.) 3√2x+3 = √y+10
Hello,

1. I presume that there must be a typo because you can't solve an equation with 2 variables.

2. Let's pretend that the equation is:

$3\sqrt{2x+3}=\sqrt{x+10}$. Now square both sides of the equation:

$9(2x+3)=x+10 \Longrightarrow 18x+27=x+10 \Longrightarrow 17x = -17 \Longrightarrow x = -1$

Now plug in this solution into the original equation and check it.

EB