03 - Solved Problems Involving Exponents and Radicals

$x^2 - 15 = -2x$

$x^2 + 2x - 15 = 0$

$(x - 3)(x + 5) = 0$

$x = 3 ~ \text{ and/or } ~ -5$
 

Check the given equation for x = 3
$\left( \dfrac{x^2 - 15}{x} \right)^2 - 16\left( \dfrac{15 - x^2}{x} \right) + 28 = 0$

$\left( \dfrac{9 - 15}{3} \right)^2 - 16\left( \dfrac{15 - 9}{3} \right) + 28 = 0$

$(-2)^2 - 16(2) + 28 = 0$

$4 - 32 + 28 = 0$   ←   okay!
 

Check the given equation for x = -5
$\left( \dfrac{x^2 - 15}{x} \right)^2 - 16\left( \dfrac{15 - x^2}{x} \right) + 28 = 0$

$\left( \dfrac{(-5)^2 - 15}{-5} \right)^2 - 16\left( \dfrac{15 - (-5)^2}{-5} \right) + 28 = 0$

$(-2)^2 - 16(2) + 28 = 0$

$4 - 32 + 28 = 0$   ←   okay!

$x^2 - 15 = -14x$

$x^2 + 14x - 15 = 0$

$(x - 1)(x + 15) = 0$

$x = 1 ~ \text{ and/or } ~ -15$
 

Check the given equation for x = 1
$\left( \dfrac{x^2 - 15}{x} \right)^2 - 16\left( \dfrac{15 - x^2}{x} \right) + 28 = 0$

$\left( \dfrac{1 - 15}{1} \right)^2 - 16\left( \dfrac{15 - 1}{1} \right) + 28 = 0$

$(-14)^2 - 16(14) + 28 = 0$

$196 - 224 + 28 = 0$   ←   okay!
 

Check the given equation for x = -15
$\left( \dfrac{x^2 - 15}{x} \right)^2 - 16\left( \dfrac{15 - x^2}{x} \right) + 28 = 0$

$\left( \dfrac{(-15)^2 - 15}{-15} \right)^2 - 16\left( \dfrac{15 - (-15)^2}{-15} \right) + 28 = 0$

$(-14)^2 - 16(14) + 28 = 0$

$196 - 224 + 28 = 0$   ←   okay!

$\dfrac{18}{\sqrt{18} + \sqrt{9 - 18}} + \dfrac{18}{\sqrt{18} - \sqrt{9 - 18}} = \dfrac{24}{\sqrt{18}}$

$\dfrac{18}{\sqrt{9(2)} + \sqrt{-9}} + \dfrac{18}{\sqrt{9(2)} - \sqrt{-9}} = \dfrac{3(8)}{\sqrt{9(2)}}$

$\dfrac{18}{3\sqrt{2} + 3i} + \dfrac{18}{3\sqrt{2} - 3i} = \dfrac{3(8)}{3\sqrt{2}}$

$\dfrac{6}{\sqrt{2} + i} + \dfrac{6}{\sqrt{2} - i} = \dfrac{8}{\sqrt{2}}$

$\dfrac{6(\sqrt{2} - i) + 6(\sqrt{2} + i)}{(\sqrt{2} + i)(\sqrt{2} - i)} = \dfrac{8}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\dfrac{6\sqrt{2} - 6i + 6\sqrt{2} + 6i}{2 - i^2} = \dfrac{8\sqrt{2}}{2}$

$\dfrac{12\sqrt{2}}{2 - (-1)} = 4\sqrt{2}$

$\dfrac{12\sqrt{2}}{3} = 4\sqrt{2}$   ←   okay!

$\dfrac{6}{\sqrt{6} + \sqrt{9 - 6}} + \dfrac{6}{\sqrt{6} - \sqrt{9 - 6}} = \dfrac{24}{\sqrt{6}}$

$\dfrac{6}{\sqrt{6} + \sqrt{3}} + \dfrac{6}{\sqrt{6} - \sqrt{3}} = \dfrac{24}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}}$

$\dfrac{6(\sqrt{6} - \sqrt{3}) + 6(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})(\sqrt{6} - \sqrt{3})} = \dfrac{24\sqrt{6}}{6}$

$\dfrac{6\sqrt{6} - 6\sqrt{3} + 6\sqrt{6} + 6\sqrt{3}}{6 - 3} = 4\sqrt{6}$

$\dfrac{12\sqrt{6}}{3} = 4\sqrt{6}$   ←   okay!