Contoh Soal Fungsi, Persamaan, dan Pertidaksamaan Kuadrat (3)

Lihat contoh soal sebelumnya di sini,  di sin, dan  di sini

\begin{array}{ll}\\ \fbox{11}.&\textrm{Sketsalah grafik fungsi kuadrat} \: \: f(x)=x^{2}-6x+5 \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|c|c|c|c|}\hline \multicolumn{5}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}:\\ &f(x)=x^{2}-6x+5\left\{\begin{matrix} x_{1}\\ \\ x_{2} \end{matrix}\right.\: \: \textrm{dan}\: \: \left\{\begin{matrix} a=1\: \: \: \: \\ b=-6\\ c=5\: \: \: \end{matrix}\right.\\ & \end{aligned}}\\\hline \textrm{Istilah}&\textrm{Memotong}&\textrm{Syarat}&\textrm{Proses}&\textrm{Titik}\\\hline \textrm{Titik potonng}&\textrm{Sumbu}-X&y=0&x^{2}-6x+5=0&\begin{aligned}&(1,0)\\ &\textrm{dan}\\ &(5,0) \end{aligned}\\\cline{2-5} &\textrm{Sumbu}-Y&x=0&y=0^{2}-6(0)+5&(0,5)\\\hline \textrm{Persamaan sumbu simetri}&\multicolumn{2}{|c|}{x=\displaystyle \frac{-b}{2a}}&\multicolumn{2}{|c|}{x=\displaystyle \frac{-(-6)}{2.1}=3}\\\hline \textrm{Koordinat titik puncak}&\multicolumn{2}{|c|}{\left ( x_{p},y_{p} \right )=\left ( \frac{-b}{2a},\frac{-D}{4a} \right )}&\begin{aligned}&\left ( \frac{-(-6)}{2.1}, \frac{-\left ( (-6)^{2}-4.1.5 \right )}{4.1}\right ) \end{aligned}&(3,-4)\\\hline \end{array}.

Untuk gambarnya perhatikanlah ilustrasi gambar di bawah ini

353

\begin{array}{ll}\\ \fbox{12}.&\textrm{Diketahu}i\: \: \alpha \: \: \textrm{dan}\: \: \beta \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\\ &x^{2}-x-2=0,\: \: \textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \alpha +\beta \: \: \textrm{dan}\: \: \alpha \beta&\textrm{e}.\quad \alpha ^{2}+\beta ^{2}\\ \textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&\textrm{f}.\quad \alpha ^{2}-\beta ^{2}\\ \textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}\\ \textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}\\ \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}-x-2=0\left\{\begin{matrix} \alpha \\ \\ \beta \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=-1\\ c=-2 \end{matrix}\right.\\ & \end{aligned}}\\\hline \begin{aligned}\textrm{a}.\quad &\alpha +\beta =-\frac{b}{a}=-\frac{(-1)}{1}=1\\ &\alpha \beta =\frac{c}{a}=\frac{(-2)}{1}=-2 \end{aligned}&\begin{aligned}\textrm{e}.\quad \alpha ^{2}+\beta ^{2}&=\left ( \alpha +\beta \right )^{2}-2\alpha \beta \\ &=1^{2}-2(-2)\\ &=1+4=5 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \left ( \alpha -\beta \right )^{2}&=\frac{D}{a^{2}}\\ &=\frac{b^{2}-4ac}{a^{2}}\\ &=\frac{(-1)^{2}-4.(1).(-2)}{(1)^{2}}\\ &=1+8=9 \end{aligned}&\begin{aligned}\textrm{f}.\quad \alpha ^{2}-\beta ^{2}&=\left ( \alpha +\beta \right )\left ( \alpha -\beta \right )\\ &=(1).(9)=9\\ &\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \displaystyle \frac{\alpha }{\beta }+\frac{\beta }{\alpha }&=\frac{\alpha ^{2}+\beta ^{2}}{\alpha \beta }\\ &=\displaystyle \frac{5}{-2}\\ &=-\frac{5}{2}\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{g}.\quad \displaystyle \frac{1}{\beta -2}+\frac{1}{\alpha -2}&=\frac{(\alpha -2)+(\beta -2)}{(\alpha -2).(\beta -2)}\\ &=\displaystyle \frac{\alpha +\beta -4}{\alpha \beta -2(\alpha +\beta )+4}\\ &=\displaystyle \frac{(-1)-4}{(-2)-2(-1)+4}\\ &=\displaystyle \frac{-5}{-2+2+4}\\ &=-\frac{5}{4} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \displaystyle \frac{1}{\beta }+\frac{1}{\alpha }&=\frac{\alpha +\beta }{\alpha \beta }\\ &=\displaystyle \frac{(-1)}{(-2)}\\ &=\frac{1}{2} \end{aligned}&\begin{aligned}\textrm{h}.\quad \displaystyle \frac{\alpha }{\beta ^{2}}+\frac{\beta }{\alpha ^{2}}&=\frac{\alpha ^{3}+\beta ^{3}}{(\alpha \beta )^{2}}\\ &=\displaystyle \frac{(\alpha +\beta )^{3}-3\alpha \beta (\alpha +\beta )}{(\alpha \beta )^{2}}\\ &=.... \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Diketahu}i\: \: p \: \: \textrm{dan}\: \: q \: \: \textrm{adalah akar-akar dari persamaan kuadrat}\\ &x^{2}+2x-5=0,\: \: \textrm{tentukanlah nilai untuk}\\ &\begin{array}{ll}\\ \textrm{a}.\quad p^{2}+q^{2}&\textrm{e}.\quad (p-3)^{2}+(q-3)^{2}\\ \textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&\textrm{f}.\quad p^{2}q+pq^{2}\\ \textrm{c}.\quad p^{3}+q^{3}&\textrm{g}.\quad (p+q)^{2}-(p-q)^{2}\\ \textrm{d}.\quad p^{3}-q^{3}&\textrm{h}.\quad (p^{3}+q^{3})-(p^{3}-q^{3})\\ \end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui}&\: \: \textrm{bahwa}\\ &x^{2}+2x-5=0\left\{\begin{matrix} p \\ \\ q \end{matrix}\right.\textrm{dan}\: \: \left\{\begin{matrix} a=1\\ b=2\\ c=-5 \end{matrix}\right.\\ & \end{aligned}}\\\hline \begin{aligned}\textrm{a}.\quad p^{2}+q^{2}&=(p+q)^{2}-2pq\\ &=(-\frac{b}{a})^{2}-2\left ( \frac{c}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{2}-2\left ( \frac{(-5)}{1} \right )\\ &=4+10\\ &=14 \end{aligned}&\begin{aligned}\textrm{e}.\quad &(p-3)^{2}+(q-3)^{2}\\ &=p^{2}-6p+9+q^{2}-6q+9\\ &=p^{2}+q^{2}-6(p+q)+18\\ &=14-6(-2)+18\\ &=14+12+18\\ &=44 \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad \displaystyle \frac{p}{q}+\frac{q}{p}&=\frac{p^{2}+q^{2}}{pq}\\ &=\displaystyle \frac{14}{-5}\\ &=-\frac{14}{5} \end{aligned}&\begin{aligned}\textrm{e}.\quad p^{2}q+pq^{2}&=pq(p+q)\\ &=(-5)(-2)\\ &=10\\ &\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad p^{3}+q^{3}&=(p+q)^{3}-3pq(p+q)\\ &=\left ( -\frac{b}{a} \right )^{3}-3\left ( \frac{c}{a} \right )\left ( -\frac{b}{a} \right )\\ &=\left ( -\frac{2}{1} \right )^{3}-3\left ( \frac{(-5)}{1} \right )\left ( -\frac{2}{1} \right )\\ &=-8-30\\ &=-38\\ & \end{aligned}&\begin{aligned}\textrm{d}.\quad p^{3}&-q^{3}\\ &=(p-q)^{3}+3pq(p-q)\\ &=\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )^{^{3}}+3\left ( \frac{c}{a} \right )\left ( \frac{\sqrt{b^{2}-4ac}}{a} \right )\\ &=\left ( \displaystyle \frac{\sqrt{2^{2}-4.1.(-5)}}{1} \right )^{3}+3\left ( \frac{-5}{1} \right )....\\ &=....\\ & \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{Untuk jawaban soal yang lain silahkan coba sendiri sebagai latihan}}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{14}&\textrm{Tentukanlah akar-akar persamaan kuadrat dengan rumus kuadrat}\\ &\begin{array}{ll}\\ \textrm{a}.\quad x^{2}-2=0&\textrm{f}.\quad 2p^{2}-5p-12=0\\ \textrm{b}.\quad x^{2}+3x-1=0&\textrm{g}.\quad 3q^{2}-11q+10=0\\ \textrm{c}.\quad x^{2}+2x-3=0&\textrm{h}.\quad 4x^{2}+11x+6=0\\ \textrm{d}.\quad x^{2}+5x-6=0&\textrm{i}.\quad 5z^{2}-z-4=0\\ \textrm{e}.\quad x^{2}-7x-8=0&\textrm{j}.\quad 6x^{2}+17x+7=0 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad x^{2}&-2=0\\ &\begin{cases} a & =1 \\ b & =0 \\ c & =-2 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x_{1,2}&=\displaystyle \frac{0\pm \sqrt{0^{2}-4.1.(-2)}}{2.1}\\ &=\displaystyle \frac{\pm \sqrt{8}}{2}=\frac{\pm \sqrt{4.2}}{2}\\ &=\displaystyle \frac{\pm 2\sqrt{2}}{2}\\ &=\pm \sqrt{2}\\ x_{1}&=\sqrt{2}\quad \textrm{atau}\quad x_{2}=-\sqrt{2}\end{aligned}&\begin{aligned}\textrm{i}.\quad 5z^{2}&-z-4=0\\ &\begin{cases} a & =5 \\ b & =-1 \\ c & =-4 \end{cases}\\ z_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ z_{1,2}&=\displaystyle \frac{-(-1)\pm \sqrt{(-1)^{2}-4.5.(-4)}}{2.5}\\ &=\displaystyle \frac{1\pm \sqrt{1+80}}{10}=\frac{1\pm \sqrt{81}}{10}\\ &=\displaystyle \frac{1\pm 9}{10}\\ z_{1}&=\displaystyle \frac{1+9}{10}=1\quad \textrm{atau}\quad z_{2}=\frac{1-9}{10}=\frac{-8}{10}=-\frac{4}{5}\\ &\end{aligned}\\\hline \end{array}.

Soal yang belum dibahas silahkan dikerjakan sendiri sebagai latihan.

\begin{array}{ll}\\ \fbox{15}&\textrm{Carilah nilai}\: \: x\: \: \textrm{yang memenuhi persamaan}\\ &\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0\end{array}\\ \textrm{Jawab}:\\ \begin{aligned}\displaystyle \frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}&=\frac{2}{x^{2}-10x-69}\\ \displaystyle \frac{1}{\left (x^{2}-10x-37 \right )+8}+\frac{1}{\left (x^{2}-10x-37 \right )-8}&=\frac{2}{\left (x^{2}-10x-37 \right )-32}\\ \textrm{Misalkan}\: \: x^{2}-10x-37=p, \: \: \textrm{maka}\qquad&\\ \displaystyle \frac{1}{p+8}+\frac{1}{p-8}&=\frac{2}{p-32}\\ \displaystyle \frac{p-8+p+8}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{2p}{(p+8)(p-8)}&=\frac{2}{p-32}\\ \displaystyle \frac{p}{p^{2}-64}&=\frac{1}{p-32}\\ p^{2}-32p&=p^{2}-64\\ p&=\displaystyle \frac{-64}{-32}\\ p&=2,\quad \textnormal{kita kembali ke bentuk semula}\\ x^{2}-10x-37&=2\\ x^{2}-10x-39&=0\\ (x-13)(x+3)&=0\\ x=13\: \: \textrm{atau}\: \: x=-3& \end{aligned}\\ \textrm{Jadi},\: \: x=13.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Tunjukkan bahwa untuk}\: \: m\in \textrm{rasional},\: \textrm{maka kedua akar persamaan}\\ &\textrm{a}.\quad x^{2}+(m+2)x+2m=0,\: \textrm{adalah rasional juga}\\ &\textrm{b}.\quad 2x^{2}+(m+4)x+(m-1)=0,\: \textrm{selalu memiliki dua akar real yang berlainan}\\ &\textrm{c}.\quad x^{2}+(m+4)x-2m^{2}-m+3=0,\: \textrm{selalu memiliki dua akar real dan rasional}\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textrm{Persamaan kuadrat}}\\\hline x^{2}+(m+2)x+2m=0&2x^{2}+(m+4)x+(m-1)=0&x^{2}+(m+4)x-2m^{2}-m+3=0\\\hline \begin{aligned}a&=1,\: b=(m+2),\: c=2m \end{aligned}&\begin{aligned}a&=2,\: b=m+4,\: c=m-1 \end{aligned}&\begin{aligned}a&=1,\: b=m+4,\: c=-2m^{2}-m+3 \end{aligned}\\\hline \multicolumn{3}{|c|}{\textrm{Jenis-jenis akar berdasarkan nilai}\: D,\: \textrm{di mana}\: D=b^{2}-4ac}\\\hline \begin{aligned}D&=(m+2)^{2}-4.1.(2m)\\ &=m^{2}+4m+4-8m\\ &=m^{2}-4m+4\\ &=(m-2)^{2} \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.2.(m-1)\\ &=m^{2}+8m+16-8m+8\\ &=m^{2}+24\\ & \end{aligned}&\begin{aligned}D&=(m+4)^{2}-4.1.(-2m^{2}-m+3)\\ &=m^{2}+8m+16+8m^{2}+4m-12\\ &=9m^{2}+12m+4\\ &=(3m+2)^{2} \end{aligned}\\\hline \multicolumn{3}{|c|}{\textrm{Kesimpulan}}\\\hline \begin{aligned}&\textrm{2 akar rasional} \end{aligned}&\begin{aligned}&\textrm{2 akar real dan berbeda} \end{aligned}&\begin{aligned}&\textrm{2 akar rasional} \end{aligned}\\\hline \end{array}.

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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