02 - Solution to Radical Equations | College Algebra Review at MATHalino

Problem 3
Find the value of $x$ from $\sqrt{(4 - x^2)^3} + 3x^2\sqrt{4 - x^2} = 0$

Solution to Problem 3

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

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

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

$\sqrt{4 - x^2} \left[ (4 - x^2) + 3x^2 \right] = 0$

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

Hence,
$\sqrt{4 - x^2} = 0 ~ \text{ and } ~ (4 + 2x^2) = 0$

For sqrt(4 - x²) = 0

$\sqrt{4 - x^2} = 0$

$4 - x^2 = 0$

$x^2 = 4$

$x = \pm 2$

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

For 4 + 2x² = 0$

$4 + 2x^2 = 0$

$2x^2 = -4$

$x^2 = -2$

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

$x = \pm \sqrt{2} \, i$

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

Thus, x = ±2 and x = ±sqrt(2)i answer

Another Solution to Problem 3

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

Let
$z = \sqrt{4 - x^2}$

$z^2 = 4 - x^2$

$x^2 = 4 - z^2$

Hence,
$z^3 + 3(4 - z^2)z = 0$

$z^3 + 12z - 3z^3 = 0$

$12z - 2z^3 = 0$

$2z(6 - z^2) = 0$

For 2z = 0; z = 0

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

$x = \pm 2$

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

For 6 - z² = 0; z = ±sqrt(6)

$x^2 = 4 - (\pm \sqrt{6})^2 = -2$

$x = \pm 2i$

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

Thus, x = ±2 and x = ±sqrt(2)i answer

Problem 4
Determine the value of $x$ from $\dfrac{1}{3x - 2} - \dfrac{8}{\sqrt{3x - 2}} = 9$

Solution to Problem 4

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

Multiply both sides of the equation by 3x - 2
$\dfrac{3x - 2}{3x - 2} - \dfrac{8(3x - 2)}{\sqrt{3x - 2}} = 9(3x - 2)$

$1 - 8\sqrt{3x - 2} = 27x - 18$

$19 - 27x = 8\sqrt{3x - 2}$

Square both sides
$(19 - 27x)^2 = 64(3x - 2)$

$361 - 1026x + 729x^2 = 192x - 128$

$729x^2 - 1218x + 489 = 0$

$243x^2 - 406x + 163 = 0$

By quadratic formula; a = 243, b = -406, and c = 163
$x = \dfrac{-(-406) \pm \sqrt{(-406)^2 - 4(243)(163)}}{2(243)}$

$x = \dfrac{406 \pm 80}{486}$

$x = 1 ~ \text{ and/or } ~ \dfrac{163}{243}$

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

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!

Thus, x = 163/243 answer

Another Solution to Problem 4

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

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

Hence,
$z^2 - 8z = 9$

$z^2 - 8z - 9 = 0$

$(z - 9)(z + 1) = 0$

$z = 9 ~ \text{ and } -1$

For z = 9

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

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

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

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

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

$x = \dfrac{163}{243}$

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!

For z = -1

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

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

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

$3x - 2 = 1$

$3x = 3$

$x = 1$

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

Thus, x = 163/243 answer

02 - Solution to Radical Equations

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