Multiply by negative three on both sides.

\(\displaystyle \frac{\sqrt{-3-7x} }{-3} \cdot -3= -9 \cdot -3\)

\(\displaystyle \sqrt{-3-7x} = 27\)

Square both sides.

\(\displaystyle (\sqrt{-3-7x}) ^2= (27)^2\)

\(\displaystyle -3-7x = 729\)

Add three on both sides.

\(\displaystyle -3-7x +3= 729+3\)

\(\displaystyle -7x=732\)

Divide by negative seven on both sides.

\(\displaystyle \frac{-7x}{-7}=\frac{732}{-7}\)

The answer is:  \(\displaystyle -\frac{732}{7}\)

Multiply the denominator on both sides to eliminate the fraction.

\(\displaystyle \frac{5}{\sqrt{5x}} \cdot \sqrt{5x}= 10 \cdot \sqrt{5x}\)

The equation becomes:

\(\displaystyle 5=10\sqrt{5x}\)

Square both sides.

\(\displaystyle 5^2=(10\sqrt{5x})^2\)

\(\displaystyle 25= 100(5x)\)

\(\displaystyle 25 =500x\)

Divide by 500 on both sides.

\(\displaystyle \frac{25}{500} =\frac{500x}{500}\)

Reduce both fractions.

The answer is:  \(\displaystyle \frac{1}{20}\)

Add 5 on both sides.  This is the same as adding \(\displaystyle \frac{25}{5}\) on both sides.

\(\displaystyle \sqrt{x}-5 +5= \frac{1}{5}+\frac{25}{5}\)

\(\displaystyle \sqrt{x} = \frac{26}{5}\)

To eliminate the radical, we must square both sides of the equation.

\(\displaystyle (\sqrt{x}) ^2=( \frac{26}{5})^2\)

The answer is:  \(\displaystyle \frac{676}{25}\)

Multiply the denominator on both sides to eliminate the denominator.

\(\displaystyle \frac{14}{\sqrt[4]{2x}+3} \cdot (\sqrt[4]{2x}+3)=4\cdot (\sqrt[4]{2x}+3)\)

Simplify both sides.

\(\displaystyle 14 = 4\sqrt[4]{2x}+12\)

Subtract 12 from both sides.

\(\displaystyle 14-12 = 4\sqrt[4]{2x}+12-12\)

The equation becomes:

\(\displaystyle 2= 4\sqrt[4]{2x}\)

Divide by 4 on both sides.

\(\displaystyle \frac{2}{4}= \frac{4\sqrt[4]{2x}}{4}\)

\(\displaystyle \frac{1}{2}=\sqrt[4]{2x}\)

Raise both sides by the power of four to eliminate the radical.

\(\displaystyle (\frac{1}{2})^4=(\sqrt[4]{2x})^4\)

\(\displaystyle \frac{1}{16} = 2x\)

Multiply by one half on both sides to isolate \(\displaystyle x\).

\(\displaystyle \frac{1}{16}\cdot \frac{1}{2} = 2x\cdot \frac{1}{2}\)

The answer is:  \(\displaystyle \frac{1}{32}\)

Subtract five from both sides of the equation.

\(\displaystyle 5-3\sqrt{9x} -5= -2-5\)

\(\displaystyle -3\sqrt{9x} = -7\)

Divide by negative three on both sides.

\(\displaystyle \frac{-3\sqrt{9x} }{-3}=\frac{ -7}{-3}\)

\(\displaystyle \sqrt{9x} = \frac{7}{3}\)

Square both sides.

\(\displaystyle (\sqrt{9x}) ^2=( \frac{7}{3})^2\)

Simplify both sides.

\(\displaystyle 9x =\frac{49}{9}\)

Multiply by one-ninth on both sides.

\(\displaystyle 9x \cdot \frac{1}{9}=\frac{49}{9}\cdot \frac{1}{9}\)

The answer is:  \(\displaystyle \frac{49}{81}\)

Cross multiply both sides.

\(\displaystyle \sqrt{2x}(\sqrt x) =(\sqrt5)(\sqrt6)\)

\(\displaystyle \sqrt{2x^2} = \sqrt{30}\)

Square both sides in order to cancel the radicals.

\(\displaystyle (\sqrt{2x^2}) ^2= (\sqrt{30})^2\)

\(\displaystyle 2x^2 = 30\)

Divide by 2 on both sides.

\(\displaystyle \frac{2x^2}{2} = \frac{30}{2}\)

\(\displaystyle x^2=15\)

Square root both sides.

\(\displaystyle \sqrt{x^2} = \sqrt{15}\)

\(\displaystyle x=\pm 15\)

Only \(\displaystyle x=15\) will satisfy the original equation.

The answer is:  \(\displaystyle 15\)

We will begin by adding 5 to both sides of the equation. 

\(\displaystyle \\\sqrt{2x+3}-5+5 = -2+5\)

\(\displaystyle \sqrt{2x+3} = 3\)

Next, let's square both sides to eliminate the radical.

\(\displaystyle (\sqrt{2x+3})^2 = 3^2\)

\(\displaystyle 2x+3 = 9\)

Finally, we can solve this like a simple two-step equation. Subtract 3 from both sides of the equation.

\(\displaystyle 2x+3-3 = 9-3\)

\(\displaystyle 2x=6\)

Now, divide each side by 2. 

\(\displaystyle \frac{2x}{2}=\frac{6}{2}\)

\(\displaystyle x=3\)

Finally, check the solution to make sure that it results in a true statement.

\(\displaystyle \sqrt{2(3)+3}=3\)

\(\displaystyle \sqrt{9}=3\)

\(\displaystyle 3=3\)

When working with radicals, a helpful step is to square both sides of an equation so that you can remove the radical sign and deal with a more classic linear or quadratic equation. But of course if you were to simply square both sides first here, you would still end up with radical signs, as were you to FOIL the left side you wouldn't eliminate the radicals.  So a good first step is to add \(\displaystyle 2\sqrt{x}\) and subtract \(\displaystyle 3\) from both sides so that you get:

\(\displaystyle x-3=2{\sqrt{x}}\)

Now when you square both sides, you'll eliminate the radical on the right-hand side and yield:

\(\displaystyle x^2-6x+9=4x\)

Then when you subtract \(\displaystyle 4x\) from both sides to set up a quadratic equalling zero, you have a factorable quadratic:

\(\displaystyle x^2-10x+9=0\)

This factors to:

\(\displaystyle (x-9)(x-1)=0\)

Which would seem to yield solutions of \(\displaystyle x=9\) and \(\displaystyle x=1\).  However, when you're solving for quadratics it's always important to plug your solutions back into the original equation to check for extraneous solutions.  Here if you plug in \(\displaystyle x=9\) the math holds: \(\displaystyle 9-2{\sqrt{9}}=3\) because \(\displaystyle 2\sqrt{9}=2(3)=6\), and \(\displaystyle 9-6=3\).

But if you plug in \(\displaystyle x=1\), you'll see that the original equation is not satisfied: \(\displaystyle 1-3\neq 2\sqrt{1}\) because \(\displaystyle -2\) does not equal \(\displaystyle 2\).  Therefore the only proper answer is \(\displaystyle x=9\).

To solve, first square both sides:

\(\displaystyle \sqrt{x-3}^2 = (x-5)^2\) squaring the left side just givs x - 3. To square the left side, use the distributite property and multiply \(\displaystyle (x-5)(x-5)\):

\(\displaystyle x - 3 = x^2 - 10x + 25\) This is a quadratic, we just need to combine like terms and get it equal to 0

\(\displaystyle 0 = x^2 -11x + 28\) now we can solve using the quadratic formula:

\(\displaystyle x = \frac{11 \pm \sqrt{11^2 - 4*1*28}}{2 } = \frac{11 \pm \sqrt {121 - 112}}{2 } = \frac{11 \pm \sqrt{9 }}{2} = \frac{11 \pm 3 }{2}\)

This gives us 2 potential answers:

\(\displaystyle x = \frac{11 + 3 }{2 } = \frac{14 }{2} = 7\) and

\(\displaystyle x = \frac{11 - 3 }{ 2 } = \frac{8}{2} = 4\)

Since the expression under the radical cannot be negative,

\(\displaystyle 4x+3\geq0\).

Solve for x:

\(\displaystyle x\geq-\frac{3}{4}\)

This is the domain, or possible \(\displaystyle x\) values, for the function.