02 - Solution to Radical Equations

$4 - x^2 = 0$

$x^2 = 4$

$x = \pm 2$
 

Check the given equation fro x = ±2
$\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$

$\sqrt{(4 - 4)^3} + 12\sqrt{4 - 4} = 0$   ←   okay

$2x^2 = -4$

$x^2 = -2$

$x = \pm \sqrt{-2}$

$x = \pm \sqrt{2} \, i$
 

Check the given equation fro x = ±sqrt(2)i
$\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$

$\sqrt{(4 - (\pm \sqrt{2}i)^2)^3} + 3(\pm \sqrt{2}i)^2\sqrt{4 - (\pm \sqrt{2}i)^2} = 0$

$\sqrt{(4 - 2i^2)^3} + 3(2i^2)\sqrt{4 - 2i^2} = 0$

$\sqrt{(4 + 2)^3} - 6\sqrt{4 + 2} = 0$

$\sqrt{6^3} - 6\sqrt{6} = 0$

$\sqrt{6^2(6) } - 6\sqrt{6} = 0$

$6\sqrt{6} - 6\sqrt{6} = 0$   ←   okay

$z^2 = 4 - x^2$

$x^2 = 4 - z^2$

$x = \pm 2$
 

Check the given equation fro x = ±2
$\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$

$\sqrt{(4 - 4)^3} + 12\sqrt{4 - 4} = 0$   ←   okay

$x = \pm 2i$
 

Check the given equation fro x = ±sqrt(2)i
$\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$

$\sqrt{(4 - (\pm \sqrt{2}i)^2)^3} + 3(\pm \sqrt{2}i)^2\sqrt{4 - (\pm \sqrt{2}i)^2} = 0$

$\sqrt{(4 - 2i^2)^3} + 3(2i^2)\sqrt{4 - 2i^2} = 0$

$\sqrt{(4 + 2)^3} - 6\sqrt{4 + 2} = 0$

$\sqrt{6^3} - 6\sqrt{6} = 0$

$\sqrt{6^2(6) } - 6\sqrt{6} = 0$

$6\sqrt{6} - 6\sqrt{6} = 0$   ←   okay

$\dfrac{1}{3 - 2} - \dfrac{8}{\sqrt{3 - 2}} = 9$

$1 - 8 \ne 9$   ←   not okay

$\dfrac{1}{3(\frac{163}{243}) - 2} - \dfrac{8}{\sqrt{3(\frac{163}{243}) - 2}} = 9$

$\dfrac{1}{\frac{1}{81}} - \dfrac{8}{\frac{1}{9}} = 9$

$81 - 72 = 9$   ←   okay!

$9 = \dfrac{1}{\sqrt{3x - 2}}$

$81 = \dfrac{1}{3x - 2}$

$3x - 2 = \dfrac{1}{81}$

$3x = \dfrac{163}{81}$

$x = \dfrac{163}{243}$
 

Check the given equation for x = 163/243
$\dfrac{1}{3x - 2} - \dfrac{8}{\sqrt{3x - 2}} = 9$

$\dfrac{1}{3(\frac{163}{243}) - 2} - \dfrac{8}{\sqrt{3(\frac{163}{243}) - 2}} = 9$

$\dfrac{1}{\frac{1}{81}} - \dfrac{8}{\frac{1}{9}} = 9$

$81 - 72 = 9$   ←   okay!

$-1 = \dfrac{1}{\sqrt{3x - 2}}$

$1 = \dfrac{1}{3x - 2}$

$3x - 2 = 1$

$3x = 3$

$x = 1$
 

Check the given equation for x = 1
$\dfrac{1}{3x - 2} - \dfrac{8}{\sqrt{3x - 2}} = 9$

$\dfrac{1}{3 - 2} - \dfrac{8}{\sqrt{3 - 2}} = 9$

$1 - 8 \ne 9$   ←   not okay