Answer Key 10.5

Terrance Berg

Answer Key 10.5

$\text{let }u=x^2$
$\therefore u^2-5u+4=0$
$\text{factors to }(u-4)(u-1)=0$
$\text{replace }u: (x^2-4)(x^2-1)=0 \\$
$(x-2)(x+2)(x-1)(x+1)=0$
$x=\pm 2, \pm 1$
$\text{let }u=y^2$
$\therefore u^2-9y+20=0$
$\text{factors to }(u-5)(u-4)=0$
$\text{replace }u: (y^2-5)(y^2-4)=0 \\$
$\begin{array}{ll} y^2-5=0\hspace{0.25in}&(y-2)(y+2)=0 \\ y^2=5&y=\pm 2 \\ y=\pm \sqrt{5}& \end{array}$
$u=m^2$
$\therefore u^2-7u-8=0$
$(u-8)(u+1)=0$
$(m^2-8)(m^2+1)=0 \\$
$(m+\sqrt{8})(m-\sqrt{8})(m^2+1)=0$
$m=\pm \sqrt{8}\text{ or }\pm 2\sqrt{2}$
$m^2+1\text{ has 2 non-real solutions}$
$u=y^2$
$\therefore u^2-29y+100=0$
$(u-25)(u-4)=0$
$(y^2-25)(y^2-4)=0 \\$
$(y-5)(y+5)(y-2)(y+2)=0$
$y=\pm 5, \pm 2$
$\text{let }u=a^2$
$\therefore u^2-50u+49=0$
$(u-49)(u-1)=0$
$(a^2-49)(a^2-1)=0 \\$
$(a-7)(a+7)(a-1)(a+1)=0$
$a=\pm 7, \pm1$
$\text{let }u=b^2$
$\therefore u^2-10u+9=0$
$(u-9)(u-1)=0$
$(b^2-9)(b^2-1)=0 \\$
$(b-3)(b+3)(b-1)(b+1)=0$
$b=\pm 3, \pm 1$
$x^4-20x^2+64=0$
$\text{let }u=x^2$
$\therefore u^2-20u+64=0$
$(u-16)(u-4)=0$
$(x^2-16)(x^2-4)=0 \\$
$(x-4)(x+4)(x-2)(x+2)=0$
$x=\pm 4, \pm 2$
$6z^6-z^3-12=0$
$\text{let }u=z^3$
$\therefore 6u^2-u-12=0$
$(3u+4)(2u-3)=0$
$(3z^3+4)(2z^3-3)=0\\$
$\begin{array}{ll} 3z^3+4=0\hspace{0.25in}&2z^3-3=0 \\ 3z^3=-4&2z^3=3 \\ \\ z^3=-\dfrac{4}{3}&z^3=\dfrac{3}{2} \\ \\ z=\sqrt[3]{-\dfrac{4}{3}}&z=\sqrt[3]{\dfrac{3}{2}} \end{array}$
$z^6-19z^3-216=0$
$\text{let }u=z^3$
$\therefore u^2-19u-216=0$
$(u-27)(u+8)=0$
$(z^3-27)(z^3+8)=0 \\$
$(z-3)(z^2+3z+9)(z+2)(z^2-2z+4)=0$
$z=3, -2$
$2\text{ non-real solutions each for the 2nd and 4th factors}$
$\text{let }u=x^3$
$\therefore u^2-35u+216=0$
$(u-27)(u-8)=0$
$(x^3-27)(x^3-8)=0$
$(x-3)(x^2+3x+9)(x-2)(x^2+2x+4)$
$x=2, 3$
$2\text{ non-real solutions each for the 2nd and 4th factors}$

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