Contoh Soal Lanjutan 4 Nilai Mutlak

$\begin{array}{ll}\\ 21.&\textrm{Jumlah akar-akar persamaan}\\ &x^{2}+\left | x \right |-6=0,\: \: \textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \color{red}\textrm{b}.&0\\ \textrm{c}.&1\\ \textrm{d}.&2\\ \textrm{e}.&4 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}x^{2}+\left | x \right |&-6=0\\ \left | x \right |^{2}+\left | x \right |&-6=0\\ \left ( \left | x \right |+3 \right )&\left ( \left | x \right |-2 \right )=0\\ \left | x \right |=-3&\: \: \textrm{atau}\: \: \left | x \right |=2\\ \textrm{tidak meme}&\textrm{nuhi (-3) atau}\: \: x=\pm 2\\ &\begin{cases} x_{1} & =2 \\ x_{2} & =-2 \end{cases}\\ x_{1}+x_{2}&=2+(-2)=0 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Penyelesaian untuk}\: \: -2\left | x-5 \right |=8\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ 0 \right \}\\ \color{red}\textrm{b}.&\left \{ \: \: \right \}\\ \textrm{c}.&\left \{ 4,5 \right \}\\ \textrm{d}.&\left \{ 1,9 \right \}\\ \textrm{e}.&\left \{ -1,9 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}-2\left | x-5 \right |&=8\\ \left | x-5 \right |&=-4\\ \textrm{Karena pe}&\textrm{rsamaan bernilai negatif},\\ \textrm{maka tida}&\textrm{k ada nilai}\: \: x\: \: \textrm{yang memenuhi}\\ \therefore \: \textbf{HP}&=\left \{ \: \: \right \} \end{aligned}\end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left |\displaystyle \frac{x-5}{2x+1} \right |=2\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-\displaystyle \frac{5}{3}\: \: \textrm{atau}\: \: \frac{3}{5}\\ \textrm{b}.&\displaystyle \frac{5}{3}\: \: \textrm{atau}\: \: -\frac{3}{5}\\ \color{red}\textrm{c}.&-\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: \frac{3}{5}\\ \textrm{d}.&\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: -\frac{3}{5}\\ \textrm{e}.&-\displaystyle \frac{7}{3}\: \: \textrm{atau}\: \: \frac{3}{7}\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\left |\displaystyle \frac{x-5}{2x+1} \right |&=2\\ \displaystyle \frac{x-5}{2x+1} &=\pm 2\\ \color{black}\textrm{untuk}&\: \: x=2\\ \displaystyle \frac{x-5}{2x+1} &=2\\ x-5&=2(2x+1)\\ x-4x&=2+5\\ -3x&=7\\ x&=\displaystyle \frac{7}{-3}=-\frac{7}{3}\\ \color{black}\textrm{untuk}&\: \: x=-2\\ x-5&=-2(2x+1)\\ x+4x&=-2+5\\ 5x&=3\\ x&=\displaystyle \frac{3}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 24.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\: \: \left |\left | 5x-4 \right |-3 \right |=2\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&-3\: \: \textrm{atau}\: \: -6\\ \color{red}\textrm{b}.&-3\: \: \textrm{atau}\: \: 6\\ \textrm{c}.&3\: \: \textrm{atau}\: \: -6\\ \textrm{d}.&3\: \: \textrm{atau}\: \: 6\\ \textrm{e}.&6\: \: \textrm{atau}\: \: 9\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\left |\left | 2x-3 \right |-4 \right |&=5\\ \left | 2x-3 \right |-4&=\pm 5\\ \left | 2x-3 \right |&=\pm 5+4\: \: \color{black}\textrm{maka},\\ \left | 2x-3 \right |&=5+4=9\: \: (\textrm{memenuhi})\\ \left | 2x-3 \right |&=-5+4=-1\\ &\color{magenta}(\textbf{tidak memenuhi})\\ \color{black}\textrm{Selanjutnya}&,\\ (2x-3)&=\pm 9\\ 2x&=\pm 9+3\\ \color{black}\textrm{untuk}&\: \: x=9\\ 2x&=9+3\\ x&=\displaystyle \frac{12}{2}=6\\ \color{black}\textrm{untuk}&\: \: x=-9\\ 2x&=-9+3\\ x&=\displaystyle \frac{-6}{2}=-3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &\textrm{persamaan}\: \: \left | x+1 \right |-2\left | x-3 \right | =5\left | x-4 \right |\: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&\displaystyle 1\frac{1}{6}\: \: \: \textrm{atau}\: \: \: 1\frac{2}{3}\\\\ \textrm{b}.&\displaystyle 1\frac{5}{6}\: \: \: \textrm{atau}\: \: \: 2\frac{2}{5}\\\\ \textrm{c}.&\displaystyle 1\frac{5}{6}\: \: \: \textrm{atau}\: \: \: 2\frac{2}{3}\\\\ \textrm{d}.&\displaystyle 2\frac{3}{4}\: \: \: \textrm{atau}\: \: \: 4\frac{2}{3}\\\\ \color{red}\textrm{e}.&\displaystyle 3\frac{1}{4}\: \: \: \textrm{atau}\: \: \: 4\frac{1}{2} \end{array}\\\\ &\textrm{Jawab}:\qquad \textrm{e}\\ &\color{blue}\begin{aligned}\left | x+1 \right |-&2\left | x-3 \right | =5\left | x-4 \right |\\ \textrm{Perhatik}&\textrm{an untuk batas sesuai definisi, maka}\\ x=-1&,\: x=3,\: \: x=4\\ \color{black}\textrm{saat}\: \: \: x&\geq 4\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=x-3,\: \: \left | x-4 \right |=x-4\\ (x+1)-&2(x-3)=5(x-4)\\ x-2x-&5x=-1-6-20\\ -6x&=-27\\ x&=\displaystyle \frac{27}{6}=4\frac{3}{6}\: \: \color{magenta}(\textrm{memenuhi}) \\ \color{black}\textrm{saat}\: \: \: 3&\leq x< 4\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=x-3,\: \: \left | x-4 \right |=4-x\\ (x+1)-&2(x-3)=5(4-x)\\ x-2x+&5x=-1-6+20\\ 4x&=13\\ x&=\displaystyle \frac{13}{4}=3\frac{1}{4}\: \: \color{magenta}(\textrm{memenuhi})\\ \color{black}\textrm{saat}\: \: \: -&1\leq x< 3\\ \left | x+1 \right |&=x+1,\: \: \left | x-3 \right |=3-x,\: \: \left | x-4 \right |=4-x\\ (x+1)-&2(3-x)=5(4-x)\\ x+2x+&5x=-1+6+20\\ 8x&=25\\ x&=\displaystyle \frac{25}{8}=3\frac{1}{8}\: \: \color{magenta}(\textrm{tidak memenuhi})\\ \color{black}\textrm{saat}\: \: \: -&1>x\\ \left | x+1 \right |&=-x-1,\: \: \left | x-3 \right |=3-x,\: \: \left | x-4 \right |=4-x\\ (-x-1)&-2(3-x)=5(4-x)\\ -x+2x&+5x=1+6+20\\ 6x&=27\\ x&=\displaystyle \frac{27}{6}=4\frac{3}{6}\: \: \color{magenta}(\textrm{tidak memenuhi}) \end{aligned} \end{array}$

Sumber Referensi

1. Kanginan, M. 2016. Matematika untuk Siswa SMA-MA/SMK-MAK Kelas X. Bandung: Srikandi Empat Widya Utama.
2. Yuana, R. A., Indriyastuti. 2017. Perspektif Matematika 1 untuk Kelas X SMA dan MA Kelompok Mata Pelajaran Wajib. Solo: PT. Tiga Serangkai Pustaka Mandiri.