Problem Solving Bentuk Bilangan Riil

Seri Pemecahan Masalah

Jika pada bahasan sebelumnya kita bahas bilangan tidak nyata atau bilangan imajiner pada akar persamaan kuadrat, sekarang kita ketengahkan bahasan sebaliknya, yaitu akar nyta atau riil dari suatu persamaan kuadrat. 

Berikut permasalahannya

(sumber soal dari blog saya sendiri di wordpress)

$\begin{array}{rl}& \text{Akar riil terbesar untuk persamaan}\\ & {\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4}\\ & \text{adalah}p+\sqrt{q+\sqrt{r}}\text{dengan}p,q,\text{dan}r\text{adalah}\\ & \text{bilangan asli}.\text{Tentukanlah nilai}p+q+r\\ \\ & \mathbf{\text{Solusi}}:\end{array}$.

$\begin{array}{rl}& {\displaystyle \frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4}\\ & \frac{3}{x-3}+1+\frac{5}{x-5}+1+\frac{17}{x-17}+1+\frac{19}{x-19}+1=x^2-11x\\ & \frac{3+(x-3)}{x-3}+\frac{5+(x-5)}{x-5}+\frac{17+(x-17)}{x-17}+\frac{19+(x-19)}{x-19}=x^2-11x\\ & \frac{x}{x-3}+\frac{x}{x-5}+\frac{x}{x-17}+\frac{x}{x-19}=x^2-11x\\ & \frac{x(x-19)+x(x-3)}{(x-3)(x-19)}+\frac{x(x-17)+x(x-5)}{(x-5)(x-17)}=x^2-11x\\ & \frac{2x^2-22x}{x^2-22x+57}+\frac{2x^2-22}{x^2-22x+85}=x^2-11x\\ & (x^2-11x)(\frac{2}{x^2-22x+57}+\frac{2}{x^2-22x+85})\\ & =x^2-11x,\text{misal}t=x^2-22x\\ & (\frac{2}{t+57}+\frac{2}{t+85})=\frac{x^2-11x}{x^2-11x}=1\\ & 2(t+85)+2(t+57)=(t+57)(t+85)\\ & 2t+170+2t+114=t^2+142t+4845\\ & 0=t^2+138t+4731\\ & t^2+138t+4731=0\{\begin{array}{c}a=1\\ b=138\\ c=4731\end{array}\\ & t_{1,2}={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}\\ & t_{1,2}={\displaystyle \frac{-138\pm \sqrt{{138}^2-4.1.4731}}{2}}\\ & ={\displaystyle \frac{-138\pm \sqrt{19044-18924}}{2}}\\ & ={\displaystyle \frac{-138\pm \sqrt{120}}{2}}\\ & ={\displaystyle \frac{-138\pm 2\sqrt{30}}{2}}\\ & =-69\pm \sqrt{30}\end{array}$.

$\begin{array}{rl}& \text{Selanjutnya}\\ & t_{1,2}=-69\pm \sqrt{30}\\ & x^2-22x=-69\pm \sqrt{30}\\ & x^2-22x+69\pm \sqrt{30}=0\\ & x^2-22x+69+\sqrt{30}=0\\ & \text{atau}{x}^2-22x+69-\sqrt{30}\\ & \\ & \text{dengan cara yang}\text{semisal diatas}\\ & \\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69+\sqrt{30})}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{{22}^2-4(69-\sqrt{30})}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{484-276-4\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{484-276+4\sqrt{30}}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm \sqrt{208-4\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm \sqrt{208+4\sqrt{30}}}{2}}\\ & x_{1,2}={\displaystyle \frac{22\pm 2\sqrt{52-\sqrt{30}}}{2}}\\ & \text{atau}{x}_{3,4}={\displaystyle \frac{22\pm 2\sqrt{52+\sqrt{30}}}{2}}\\ & x_{1,2}=11\pm \sqrt{52-\sqrt{30}}\\ & \text{atau}{x}_{3,4}=11\pm \sqrt{52+\sqrt{30}}\\ & \\ & \text{Maka},\\ & \{\begin{array}{c}{x}_1=11+\sqrt{52-\sqrt{30}}\\ \\ x_2=11-\sqrt{52-\sqrt{30}}\end{array}\\ & \text{atau}\{\begin{array}{c}{x}_3=11+\sqrt{52+\sqrt{30}}\\ \\ x_4=11-\sqrt{52+\sqrt{30}}\end{array}\end{array}$.

$\begin{array}{rl}\text{Selanjutnya nilai}& \text{yang paling pas sesuai soal adalah}:\\ & x_3=11+\sqrt{52+\sqrt{30}}=p+\sqrt{q+\sqrt{r}}\\ \text{Sehingga nilai}& p+q+r=11+52+30=93\end{array}$.