01 - Solution to Radical Equations

$\sqrt{3 - x} + \sqrt{4 - 2x} = \sqrt{3 - 3x}$

$\left( \sqrt{3 - x} + \sqrt{4 - 2x} \right)^2 = \left( \sqrt{3 - 3x} \right)^2$

$(3 - x) + 2\sqrt{3 - x} \cdot \sqrt{4 - 2x} + (4 - 2x) = (3 - 3x)$

$2\sqrt{3 - x} \cdot \sqrt{4 - 2x} = -4$

$\sqrt{3 - x} \cdot \sqrt{4 - 2x} = -2$

$\left( \sqrt{3 - x} \cdot \sqrt{4 - 2x} \right)^2 = (-2)^2$

$(3 - x)(4 - 2x) = 4$

$12 - 10x + 2x^2 = 4$

$2x^2 - 10x + 8 = 0$

$x^2 - 5x + 4 = 0$

$(x - 4)(x - 1) = 0$

$x = 4 \, \text{ and/or } \, 1$
 

For x = 4
$\sqrt{3 - x} + \sqrt{4 - 2x} = \sqrt{3 - 3x}$

$\sqrt{3 - 4} + \sqrt{4 - 8} = \sqrt{3 - 12}$

$\sqrt{-1} + \sqrt{-4} = \sqrt{-9}$

$i + 2i = 3i$

$3i = 3i$   ←   okay
 

For x = 1
$\sqrt{3 - x} + \sqrt{4 - 2x} = \sqrt{3 - 3x}$

$\sqrt{3 - 1} + \sqrt{4 - 2} = \sqrt{3 - 3}$

$\sqrt{2} + \sqrt{2} \ne 0$   ←   not okay
 

Thus, x = 4           answer

$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

$\dfrac{\sqrt{2x + 4}}{\sqrt{x - 5}} + \dfrac{8\sqrt{x - 5}}{\sqrt{2x + 4}} = 6$

$\dfrac{\left( \sqrt{2x + 4} \right)^2 + 8 \left( \sqrt{x - 5} \right)^2}{\sqrt{x - 5} \, \sqrt{2x + 4}} = 6$

$\dfrac{(2x + 4) + 8(x - 5)}{\sqrt{x - 5} \, \sqrt{2x + 4}} = 6$

$\dfrac{2x + 4 + 8x - 40}{\sqrt{x - 5} \, \sqrt{2x + 4}} = 6$

$\dfrac{10x - 36}{\sqrt{x - 5} \, \sqrt{2x + 4}} = 6$

$10x - 36 = 6\sqrt{x - 5} \, \sqrt{2x + 4}$

$5x - 18 = 3\sqrt{x - 5} \, \sqrt{2x + 4}$

$(5x - 18)^2 = \left( 3\sqrt{x - 5} \, \sqrt{2x + 4} \right)^2$

$25x^2 - 180x + 324 = 9(x - 5)(2x + 4)$

$25x^2 - 180x + 324 = 9(2x^2 - 6x - 20)$

$25x^2 - 180x + 324 = 18x^2 - 54x - 180$

$7x^2 - 126x + 504 = 0$

$x^2 - 18x + 72 = 0$

$(x - 12)(x - 6) = 0$

$x = 12 \, \text{ and/or } \, 6$
 

For x = 12
$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

$\sqrt{\dfrac{24 + 4}{12 - 5}} + 8\sqrt{\dfrac{12 - 5}{24 + 4}} = 6$

$\sqrt{\dfrac{28}{7}} + 8\sqrt{\dfrac{7}{8}} = 6$   ←   okay!
 

For x = 6
$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

$\sqrt{\dfrac{12 + 4}{6 - 5}} + 8\sqrt{\dfrac{6 - 5}{12 + 4}} = 6$

$\sqrt{\dfrac{16}{1}} + 8\sqrt{\dfrac{1}{16}} = 6$   ←   okay!
 

Thus, x = 12 and x = 6           answer

$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$
 

Let
$z = \sqrt{\dfrac{2x + 4}{x - 5}}$

Thus,
$z + 8\left( \dfrac{1}{z} \right) = 6$

$z^2 + 8 = 6z$

$z^2 - 6z + 8 = 0$

$(z - 4)(z - 2) = 0$

$z = 4 \, \text{ and } \, 2$
 

For z = 4

$z = \sqrt{\dfrac{2x + 4}{x - 5}}$

$4 = \sqrt{\dfrac{2x + 4}{x - 5}}$

$16 = \dfrac{2x + 4}{x - 5}$

$16(x - 5) = 2x + 4$

$16x - 80 = 2x + 4$

$14x = 84$

$x = 6$
 

Check the given equation for x = 6
$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

$\sqrt{\dfrac{12 + 4}{6 - 5}} + 8\sqrt{\dfrac{6 - 5}{12 + 4}} = 6$

$\sqrt{\dfrac{16}{1}} + 8\sqrt{\dfrac{1}{16}} = 6$   ←   okay!

 

For z = 2

$z = \sqrt{\dfrac{2x + 4}{x - 5}}$

$2 = \sqrt{\dfrac{2x + 4}{x - 5}}$

$4 = \dfrac{2x + 4}{x - 5}$

$4(x - 5) = 2x + 4$

$4x - 20 = 2x + 4$

$2x = 24$

$x = 12$
 

Check the given equation for x = 12
$\sqrt{\dfrac{2x + 4}{x - 5}} + 8\sqrt{\dfrac{x - 5}{2x + 4}} = 6$

$\sqrt{\dfrac{24 + 4}{12 - 5}} + 8\sqrt{\dfrac{12 - 5}{24 + 4}} = 6$

$\sqrt{\dfrac{28}{7}} + 8\sqrt{\dfrac{7}{28}} = 6$   ←   okay!

 

Thus, x = 12 and x = 6           answer