Answer Key 10.5

Terrance Berg

Answer Key 10.5

$let u = x^{2}$
$∴ u^{2} - 5 u + 4 = 0$
$factors to (u - 4) (u - 1) = 0$
$replace u : (x^{2} - 4) (x^{2} - 1) = 0$
$(x - 2) (x + 2) (x - 1) (x + 1) = 0$
$x = \pm 2, \pm 1$
$let u = y^{2}$
$∴ u^{2} - 9 y + 20 = 0$
$factors to (u - 5) (u - 4) = 0$
$replace u : (y^{2} - 5) (y^{2} - 4) = 0$
$\begin{array}{ll} y^{2} - 5 = 0 & (y - 2) (y + 2) = 0 \\ y^{2} = 5 & y = \pm 2 \\ y = \pm \sqrt{5} \end{array}$
$u = m^{2}$
$∴ u^{2} - 7 u - 8 = 0$
$(u - 8) (u + 1) = 0$
$(m^{2} - 8) (m^{2} + 1) = 0$
$(m + \sqrt{8}) (m - \sqrt{8}) (m^{2} + 1) = 0$
$m = \pm \sqrt{8} or \pm 2 \sqrt{2}$
$m^{2} + 1 has 2 non-real solutions$
$u = y^{2}$
$∴ u^{2} - 29 y + 100 = 0$
$(u - 25) (u - 4) = 0$
$(y^{2} - 25) (y^{2} - 4) = 0$
$(y - 5) (y + 5) (y - 2) (y + 2) = 0$
$y = \pm 5, \pm 2$
$let u = a^{2}$
$∴ u^{2} - 50 u + 49 = 0$
$(u - 49) (u - 1) = 0$
$(a^{2} - 49) (a^{2} - 1) = 0$
$(a - 7) (a + 7) (a - 1) (a + 1) = 0$
$a = \pm 7, \pm 1$
$let u = b^{2}$
$∴ u^{2} - 10 u + 9 = 0$
$(u - 9) (u - 1) = 0$
$(b^{2} - 9) (b^{2} - 1) = 0$
$(b - 3) (b + 3) (b - 1) (b + 1) = 0$
$b = \pm 3, \pm 1$
$x^{4} - 20 x^{2} + 64 = 0$
$let u = x^{2}$
$∴ u^{2} - 20 u + 64 = 0$
$(u - 16) (u - 4) = 0$
$(x^{2} - 16) (x^{2} - 4) = 0$
$(x - 4) (x + 4) (x - 2) (x + 2) = 0$
$x = \pm 4, \pm 2$
$6 z^{6} - z^{3} - 12 = 0$
$let u = z^{3}$
$∴ 6 u^{2} - u - 12 = 0$
$(3 u + 4) (2 u - 3) = 0$
$(3 z^{3} + 4) (2 z^{3} - 3) = 0$
$\begin{array}{ll} 3 z^{3} + 4 = 0 & 2 z^{3} - 3 = 0 \\ 3 z^{3} = - 4 & 2 z^{3} = 3 \\ z^{3} = - \frac{4}{3} & z^{3} = \frac{3}{2} \\ z = \sqrt[3]{- \frac{4}{3}} & z = \sqrt[3]{\frac{3}{2}} \end{array}$
$z^{6} - 19 z^{3} - 216 = 0$
$let u = z^{3}$
$∴ u^{2} - 19 u - 216 = 0$
$(u - 27) (u + 8) = 0$
$(z^{3} - 27) (z^{3} + 8) = 0$
$(z - 3) (z^{2} + 3 z + 9) (z + 2) (z^{2} - 2 z + 4) = 0$
$z = 3, - 2$
$2 non-real solutions each for the 2nd and 4th factors$
$let u = x^{3}$
$∴ u^{2} - 35 u + 216 = 0$
$(u - 27) (u - 8) = 0$
$(x^{3} - 27) (x^{3} - 8) = 0$
$(x - 3) (x^{2} + 3 x + 9) (x - 2) (x^{2} + 2 x + 4)$
$x = 2, 3$
$2 non-real solutions each for the 2nd and 4th factors$

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