Please, see attachment for quadratic equation. As a first step, do I rewrite sqrt{22x} as (22x)^(1/2)?

Is this correct: x^2 + (22x)^(1/2) = 0

See attachment.

2. Originally Posted by sologuitar
Please, see attachment for quadratic equation. As a first step, do I rewrite sqrt{22x} as (22x)^(1/2)?

Is this correct: x^2 + (22x)^(1/2) = 0

See attachment.

3. Originally Posted by sologuitar
Please, see attachment for quadratic equation. As a first step, do I rewrite sqrt{22x} as (22x)^(1/2)?

Is this correct: x^2 + (22x)^(1/2) = 0

See attachment.
If you like, it doesn't make a difference

x^2 = -\sqrt{22x}

Square both sides

$x^4 = 22x$

$x^4-22x=0$

$x(x^3-22)=0$

Either $x=0$ or $x^3-22=0$

You can use the difference of two cubes to solve

I get
$x=0$

$x = \sqrt [3]{22}$

$x=\frac{-\sqrt [3]{22}+i\,\sqrt{3\sqrt [3]{22^2}}}{2}$

$x=\frac{-\sqrt [3]{22}-i\,\sqrt{3\sqrt [3]{22^2}}}{2}$

Some of these answers may not be correct, sub into the original equation to check. Also the last two answers are complex

4. ## OK...

Originally Posted by e^(i*pi)
If you like, it doesn't make a difference

x^2 = -\sqrt{22x}

Square both sides

$x^4 = 22x$

$x^4-22x=0$

$x(x^3-22)=0$

Either $x=0$ or $x^3-22=0$

You can use the difference of two cubes to solve

I get
$x=0$

$x = \sqrt [3]{22}$

$x=\frac{-\sqrt [3]{22}+i\,\sqrt{3\sqrt [3]{22^2}}}{2}$

$x=\frac{-\sqrt [3]{22}-i\,\sqrt{3\sqrt [3]{22^2}}}{2}$

Some of these answers may not be correct, sub into the original equation to check. Also the last two answers are complex