$\begin{aligned}x_{1}+x_{2} & =-\frac{b}{a}\\
& =-\frac{(-1)}{2}\\
& =\frac{1}{2}
\end{aligned}
$

Dari persamaan kuadrat $x^{2}-3x+4=0$ didapat:

$\alpha+\beta=3$ dan $\alpha\beta=4$

Jadi $\begin{aligned}\frac{1}{\alpha}+\frac{1}{\beta} & =\frac{\alpha+\beta}{\alpha\beta}\\
& =\frac{3}{4}
\end{aligned}
$

Dari persamaan kuadrat $2x^{2}+6x-2=0$ didapat :

$\alpha+\beta=-3$ dan $\alpha\beta=-1$

Jadi $\begin{aligned}\alpha^{2}+\beta^{2} & =\left(\alpha+\beta\right)^{2}-2\alpha\beta\\
& =\left(-3\right)^{2}-2(-1)\\
& =9+2\\
& =11
\end{aligned}
$

Dari persamaan kuadrat $2x^{2}-4x+5=0$, diperoleh :

$\alpha+\beta=2$ dan $\alpha\beta=\frac{5}{2}$

$\begin{aligned}\alpha^{3}\beta+\beta^{3}\alpha & =\alpha\beta\left(\alpha^{2}+\beta^{2}\right)\\
& =\alpha\beta\left[\left(\alpha+\beta\right)^{2}-2\alpha\beta\right]\\
& =\left(\frac{5}{2}\right)\left[\left(2\right)^{2}-2\left(\frac{5}{2}\right)\right]\\
& =\frac{5}{2}\left(4-5\right)\\
& =-\frac{5}{2}
\end{aligned}
$

Dari persamaan kuadrat $x^{2}+(m-3)x+m=0$ diperoleh :

$\alpha+\beta=3-m$ dan $\alpha\beta=m$

$\frac{1}{\alpha}+\frac{1}{\beta}=2\rightarrow\frac{\alpha+\beta}{\alpha\beta}=2$

$=\frac{3-m}{m}=2\rightarrow3-m=2m\rightarrow m=1$

Dari persamaan $x^{2}-2x+2=0$ dapat diperoleh :

$\alpha+\beta=2$ dan $\alpha\beta=2$

$\alpha\left(\alpha\beta^{-1}-\beta\right)+\beta\left(\beta\alpha^{-1}-\alpha\right)=\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}-2\alpha\beta$

$=\frac{\alpha^{3}+\beta^{3}}{\alpha\beta}-2\alpha\beta$

$=\frac{\left(\alpha+\beta\right)^{3}-3\alpha\beta\left(\alpha+\beta\right)}{\alpha\beta}-2\alpha\beta$

$=\frac{\left(2\right)^{3}-3(2)(2)}{2}-2(2)$

$=-2-4=-6$

Misal persamaan kuadrat $x^{2}-3x+n=0$ akar-akarnya $\alpha$ dan $\beta$, maka diperoleh :

$\alpha+\beta=3$ dan $\alpha\beta=n$

Misal persamaan kuadrat $x^{2}+x-n=0$ akar-akarnya $u$ dan $v$, maka diperoleh :

$u+v=-1$ dan $uv=-n$

$\alpha^{2}+\beta^{2}=u^{3}+v^{3}$

$\left(\alpha+\beta\right)^{2}-2\alpha\beta=\left(u+v\right)^{3}-3uv\left(u+v\right)$

$\left(3\right)^{2}-2n=(-1)^{3}-3(-n)(-1)$

$9-2n=-1-3n$

$n=-10$

$\log\left(64\cdot\sqrt[24]{2^{x^{2}-40x}}\right)=\log\left(2^{6}\cdot2^{\frac{x^{2}-40x}{24}}\right)=0$

$\log\left(2^{6}\cdot2^{\frac{x^{2}-40x}{24}}\right)=log1$

$\left(2^{6+\frac{x^{2}-40x}{24}}\right)=2^{0}\rightarrow6+\frac{x^{2}-40x}{24}=0$

$\frac{x^{2}-40x+144}{24}=0\rightarrow x^{2}-40x+144=0$

Jadi $x_{1}x_{2}=144$

$x_{1}$ dan $x_{2}$ akar-akar dari $x^{2}-ax+6=0$ diperoleh :

$x_{1}+x_{2}=a$ dan $x_{1}x_{2}=6$

$y_{1}$dan $y_{2}$ akar-akar dari $4x^{2}-(a+3)x+3=0$ diperoleh :

$y_{1}+y_{2}=\frac{a+3}{4}$ dan $y_{1}y_{2}=\frac{3}{4}$

Salah satu akar persamaan kuadrat $x^{2}-ax+6=0$ adalah kebalikan dari salah satu akar dari persamaan $4x^{2}-(a+3)x+3=0$, maka $x_{1}=\frac{1}{y_{1}}$

$*\ x_{1}x_{2}=6\rightarrow\frac{1}{y_{1}}x_{2}=6\rightarrow y_{1}=\frac{x_{2}}{6}$

$*\ y_{1}y_{2}=\frac{3}{4}\rightarrow\frac{1}{x_{1}}y_{2}=\frac{3}{4}\rightarrow y_{2}=\frac{3}{4}x_{1}$

$*\ y_{1}+y_{2}=\frac{a+3}{4}$

$\frac{x_{2}}{6}+\frac{3}{4}x_{1}=\frac{a+3}{4}\rightarrow2x_{2}+9x_{1}=3a+9$…..pers(1)

$*\ x_{1}+x_{2}=a\rightarrow2x_{1}+2x_{2}=2a$……pers (2)

Dari persamaan (1) dan (2) diperoleh $a=5$ atau $a=-\frac{25}{2}$

Salah satu akarnya$\alpha=\sqrt[4]{31-8\sqrt{15}}$ dan akar lainnya dimisalkan $\beta$

$\begin{aligned}\alpha & =\sqrt[4]{31-8\sqrt{15}}\\
& =\sqrt{\sqrt{31-2\sqrt{4^{2}\cdot15}}}\\
& =\sqrt{4-\sqrt{15}}\\
& =\frac{1}{\sqrt{2}}\left(\sqrt{8-2\sqrt{15}}\right)\\
& =\frac{1}{\sqrt{2}}\left(\sqrt{5}-\sqrt{3}\right)\\
& =\frac{1}{2}\left(\sqrt{10}-\sqrt{6}\right)
\end{aligned}
$

$*\ \alpha+\beta=\sqrt{10}$

$\begin{aligned}\beta & =\sqrt{10}-\alpha\\
& =\sqrt{10}-\frac{1}{2}\sqrt{10}+\frac{1}{2}\sqrt{6}\\
& =\frac{1}{2}\sqrt{10}+\frac{1}{2}\sqrt{6}
\end{aligned}
$

$*\ \alpha\beta=p$

$\begin{aligned}p & =\left(\frac{1}{2}\sqrt{10}-\frac{1}{2}\sqrt{6}\right)\left(\frac{1}{2}\sqrt{10}+\frac{1}{2}\sqrt{6}\right)\\
& =\frac{10}{4}-\frac{6}{4}\\
& =1
\end{aligned}
$

Misal $y_{1}$ dan $y_{2}$ adalah akar-akar persamaan $y^{2}-2y+a=0$, sehingga diperoleh :

$y_{1}+$$y_{2}=2$ dan $y_{1}$$y_{2}=a$

Misal $x_{1}$ dan $x_{2}$ adalah akar-akar persamaan $x^{2}-bx-32=0$, sehingga diperoleh :

$x_{1}+$$x_{2}=b$ dan $x_{1}$$x_{2}=-32$

$y_{1}=x_{1}+3$ dan $y_{2}=x_{2}+3$

(i) $y_{1}+$$y_{2}=x_{1}+x_{2}+6$

$2=b+6\rightarrow b=-4$

(ii) $\begin{aligned}y_{1}y_{2} & =\left(x_{1}+3\right)\left(x_{2}+3\right)\\
& =x_{1}x_{2}+3\left(x_{1}+x_{2}\right)+9
\end{aligned}
$

$a=-32+3(-4)+9=-35$

Jadi $a+b=-35-4=-39$

Misal akarnya adalah $\alpha=\sqrt{21+4\sqrt{5}}$ dan $\beta$

$\begin{aligned}\alpha & =\sqrt{21+4\sqrt{5}}\\
& =\sqrt{21+2\sqrt{20}}\\
& =\sqrt{20}+1
\end{aligned}
$

(i)$\alpha\beta=\frac{38}{2}=19$

$\left(\sqrt{20}+1\right)\beta=19$

$\beta=\frac{19}{\sqrt{20}+1}$

(ii)$\alpha+\beta=\frac{a}{2}$

$\sqrt{20}+1+\frac{19}{\sqrt{20}+1}=\frac{a}{2}$

$\frac{40+4\sqrt{5}}{\sqrt{20}+1}=\frac{a}{2}$

$\begin{aligned}\frac{a}{2} & =\frac{40+4\sqrt{5}}{\sqrt{20}+1}\times\frac{\sqrt{20}-1}{\sqrt{20}-1}\\
& =\frac{76}{19}\sqrt{5}\\
& =4\sqrt{5}
\end{aligned}
$

$a=8\sqrt{5}$

$x^{2}-px+q=0\rightarrow\alpha+\beta=p$ dan $\alpha\beta=q$

$2x^{2}-(p+3)x+4(q-8)=0$ akar-akarnya $\left(\alpha-1\right)$ dan $\left(\beta-1\right)$ sehingga diperoleh :

(i)$\begin{aligned}\left(\alpha-1\right)+\left(\beta-1\right) & =\left(\alpha+\beta\right)-2\\
& =\frac{p+3}{2}
\end{aligned}
$

$p-2=\frac{p+3}{2}$

$2p-4=p+3\rightarrow p=7$

(ii)$\begin{aligned}\left(\alpha-1\right)\left(\beta-1\right) & =\alpha\beta-\left(\alpha+\beta\right)+1\\
& =2\left(q-8\right)
\end{aligned}
$

$q-p+1=2q-16$

$q-7+1=2q-16\rightarrow q=10$

Jadi $p-q=7-10=-3$

Dari persamaan $x^{2}+ax+b=0$ diperoleh :

$\alpha+\beta=-a$ dan $\alpha\beta=b$

Dari persamaan $x^{2}+\left(a+2b-2\right)x+\left(b+2a+2\right)=0$ diperoleh :

$\alpha^{2}+\beta^{2}=-a-2b+2$dan $\alpha^{2}\beta^{2}=b+2a+2$

(i) $\alpha^{2}+\beta^{2}=a+2b-2$

$\left(\alpha+\beta\right)^{2}-2\alpha\beta=a+2b-2$

$\left(-a\right)^{2}-2b=-a-2b+2$

$a^{2}+a-2=0$

$\left(a+1\right)\left(a-2\right)=0$

$a=-1$ atau $a=2$

Karena $a$ anggota bilangan asli, maka yang memenuhi adalah $a=2$

(ii) $\alpha^{2}\beta^{2}=b+2a+2$

$\left(\alpha\beta\right)^{2}=b+2(2)+2$

$b^{2}-b-6=0$

$\left(b-3\right)\left(b+2\right)=0$

$b=3$ atau $b=-2$

Karena $b$ anggota bilangan asli, maka yang memenuhi adalah $b=3$

Jadi $a+b=5$

Dari persamaan kuadrat $4x^{2}+px+4=0$ dapat diperoleh :

$\alpha+\beta=-\frac{p}{4}$ dan $\alpha\beta=1$

$\frac{\frac{1}{\alpha}+\frac{1}{\beta}}{\alpha^{3}+\beta^{3}}=16$

$\frac{\frac{\alpha+\beta}{\alpha\beta}}{\left(\alpha+\beta\right)^{3}-3\alpha\beta\left(\alpha+\beta\right)}=16$

$\frac{-\frac{p}{4}}{\left(-\frac{p}{4}\right)^{3}-3\left(-\frac{p}{4}\right)}=16$

Misal $y=$-$\frac{p}{4}$, sehingga persamaan diatas menjadi :

$\frac{y}{y^{3}-3y}=16\rightarrow y=16y^{3}-48y$

$16y^{3}-49y=0$

$y\left(16y^{2}-49\right)=y\left(4y+7\right)\left(4y-7\right)=0$

$y=0$ atau $y=-\frac{7}{4}$ atau $y=\frac{7}{4}$

$*\ y=0\rightarrow-\frac{p}{4}=0\rightarrow p=0$

$*\ y=-\frac{7}{4}\rightarrow-\frac{p}{4}=-\frac{7}{4}\rightarrow p=7$

$*\ y=\frac{7}{4}\rightarrow-\frac{p}{4}=\frac{7}{4}\rightarrow p=-7$

Jadi nilai $p$ yang memenuhi adalah $-7$ dan $7$