HOTS Zone : Persamaan Kuadrat [2]
Table of Contents
Tipe:
No.
Tinjau persamaan yang berbentukALTERNATIF PENYELESAIAN
\(\begin{aligned}
D&\ge 0\\
b^2-4c&\ge 0\\
4c&\le b^2\\
c&\le\dfrac{b^2}4
\end{aligned}\)
Kita lihat bahwa nilai c terbesar adalah nilai terkecil antara 6, atau bilangan bulat terbesar yang kurang dari atau sama dengan $\dfrac{b^2}4$. Jadi mencari banyaknya c yang memenuhi sama dengan mencari $\min\left(6,\left\lfloor\dfrac{b^2}4\right\rfloor\right)$. Untuk1 ≤ b ≤ 6 ,
\(\begin{aligned} \displaystyle\sum_{b=1}^6\min\left(6,\left\lfloor\dfrac{b^2}4\right\rfloor\right)&=\min\left(6,\left\lfloor\dfrac{1^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{2^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{3^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{4^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{5^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{6^2}4\right\rfloor\right)\\[4pt] &=\min\left(6,0\right)+\min\left(6,1\right)+\min\left(6,2\right)+\min\left(6,4\right)+\min\left(6,6\right)+\min\left(6,9\right)\\[4pt] &=0+1+2+4+6+6\\ &=\color{blue}\boxed{\boxed{\color{black}19}} \end{aligned}\)
Kita lihat bahwa nilai c terbesar adalah nilai terkecil antara 6, atau bilangan bulat terbesar yang kurang dari atau sama dengan $\dfrac{b^2}4$. Jadi mencari banyaknya c yang memenuhi sama dengan mencari $\min\left(6,\left\lfloor\dfrac{b^2}4\right\rfloor\right)$. Untuk
\(\begin{aligned} \displaystyle\sum_{b=1}^6\min\left(6,\left\lfloor\dfrac{b^2}4\right\rfloor\right)&=\min\left(6,\left\lfloor\dfrac{1^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{2^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{3^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{4^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{5^2}4\right\rfloor\right)+\min\left(6,\left\lfloor\dfrac{6^2}4\right\rfloor\right)\\[4pt] &=\min\left(6,0\right)+\min\left(6,1\right)+\min\left(6,2\right)+\min\left(6,4\right)+\min\left(6,6\right)+\min\left(6,9\right)\\[4pt] &=0+1+2+4+6+6\\ &=\color{blue}\boxed{\boxed{\color{black}19}} \end{aligned}\)
Jadi, ada 19 persamaan demikian yang memiliki akar-akar real jika koefisien b dan c hanya boleh dipilih dari himpunan {1, 2, 3, 4, 5, 6} .
No.
Misalkan αn dan βn merupakan akar-akar dari persamaan- $-\dfrac1{43}$
- $-\dfrac{43}{47}$
- $-\dfrac{46}{2021}$
- $-\dfrac{2020}{2021}$
- −1
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac1{\left(\alpha_{n}+2\right)\left(\beta_{n}+2\right)}&=\dfrac1{\alpha_{n}\beta_{n}+2\left(\alpha_{n}+\beta_{n}\right)+4}\\[4pt]
&=\dfrac1{\dfrac{-2n^2}2+2\left(\dfrac{-(n+4)}2\right)+4}\\[4pt]
&=\dfrac1{-n^2-n-4+4}\\[4pt]
&=\dfrac1{-n^2-n}\\[4pt]
&=\dfrac1{n+1}-\dfrac1n
\end{aligned}\)
\(\begin{aligned} \dfrac1{\left(\alpha_{43}+2\right)\left(\beta_{43}+2\right)}+\dfrac1{\left(\alpha_{44}+2\right)\left(\beta_{44}+2\right)}+\cdots+\dfrac1{\left(\alpha_{2020}+2\right)\left(\beta_{2020}+2\right)}&=\dfrac1{44}-\dfrac1{43}+\dfrac1{45}-\dfrac1{44}+\cdots+\dfrac1{2021}-\dfrac1{2020}\\[4pt] &=-\dfrac1{43}+\dfrac1{2021}\\ &=\color{blue}\boxed{\boxed{\color{black}-\dfrac{46}{2021}}} \end{aligned}\)
\(\begin{aligned} \dfrac1{\left(\alpha_{43}+2\right)\left(\beta_{43}+2\right)}+\dfrac1{\left(\alpha_{44}+2\right)\left(\beta_{44}+2\right)}+\cdots+\dfrac1{\left(\alpha_{2020}+2\right)\left(\beta_{2020}+2\right)}&=\dfrac1{44}-\dfrac1{43}+\dfrac1{45}-\dfrac1{44}+\cdots+\dfrac1{2021}-\dfrac1{2020}\\[4pt] &=-\dfrac1{43}+\dfrac1{2021}\\ &=\color{blue}\boxed{\boxed{\color{black}-\dfrac{46}{2021}}} \end{aligned}\)
Jadi, $\dfrac1{\left(\alpha_{43}+2\right)\left(\beta_{43}+2\right)}+\dfrac1{\left(\alpha_{44}+2\right)\left(\beta_{44}+2\right)}+\cdots+\dfrac1{\left(\alpha_{2020}+2\right)\left(\beta_{2020}+2\right)}=-\dfrac{46}{2021}$.
JAWAB: C
JAWAB: C
No.
Himpunan S adalah himpunan semua bilangan bulat x yang mengakibatkanALTERNATIF PENYELESAIAN
Misal x2 − 5x − 1 = p2
\(\begin{aligned} x^2-5x-1-p^2&=0\\ x&=\dfrac{5\pm\sqrt{(-5)^2-4(-1-p^2)}}2\\[4pt] &=\dfrac{5\pm\sqrt{25+4+4p^2}}2\\[4pt] &=\dfrac{5\pm\sqrt{29+4p^2}}2 \end{aligned}\)
Kita lihat bahwa $\sqrt{29+4p^2}$ harus berupa bilangan bulat. Misal $\sqrt{29+4p^2}=q$
\(\begin{aligned} 29+4p^2&=q^2\\ 29&=q^2-4p^2\\ 29&=(q+2p)(q-2p)\\ \end{aligned}\)
Karena 29 merupakan bilangan prima, didapat
\(\begin{aligned} q+2p&=29\\ q-2p&=1&-\\\hline 4p&=28\\ p&=7\\ p^2=49 \end{aligned}\)
\(\begin{aligned} x^2-5x-1&=49\\ x^2-5x-50&=0 \end{aligned}\)
Hasil kali akar-akarnya adalah −50.
\(\begin{aligned} x^2-5x-1-p^2&=0\\ x&=\dfrac{5\pm\sqrt{(-5)^2-4(-1-p^2)}}2\\[4pt] &=\dfrac{5\pm\sqrt{25+4+4p^2}}2\\[4pt] &=\dfrac{5\pm\sqrt{29+4p^2}}2 \end{aligned}\)
Kita lihat bahwa $\sqrt{29+4p^2}$ harus berupa bilangan bulat. Misal $\sqrt{29+4p^2}=q$
\(\begin{aligned} 29+4p^2&=q^2\\ 29&=q^2-4p^2\\ 29&=(q+2p)(q-2p)\\ \end{aligned}\)
Karena 29 merupakan bilangan prima, didapat
\(\begin{aligned} q+2p&=29\\ q-2p&=1&-\\\hline 4p&=28\\ p&=7\\ p^2=49 \end{aligned}\)
\(\begin{aligned} x^2-5x-1&=49\\ x^2-5x-50&=0 \end{aligned}\)
Hasil kali akar-akarnya adalah −50.
Jadi, hasil kali semua anggota S adalah −50.
JAWAB: C
JAWAB: C
No.
Diketahui a adalah solusi dari persamaan- 0
- −2
- 2
- 1
- −1
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \dfrac{2a^5-5a^4+2a^3-8a^2}{a^2+1}&=\dfrac{2a^5-6a^4+2a^3+a^4-3a^3+a^2+3a^3-9a^2+3a-3a}{a^2-3a+1+3a}\\[4pt] &=\dfrac{2a^3\left(a^2-3a+1\right)+a^2\left(a^2-3a+1\right)+3a\left(a^2-3a+1\right)-3a}{a^2-3a+1+3a}\\[4pt] &=\dfrac{2a^3\left(0\right)+a^2\left(0\right)+3a\left(0\right)-3a}{0+3a}\\[4pt] &=\dfrac{-3a}{3a}\\ \color{blue}\boxed{\boxed{\color{black}-1}} \end{aligned}\)
Jadi, \(\dfrac{2a^5-5a^4+2a^3-8a^2}{a^2+1}=-1\).
JAWAB: E
JAWAB: E
Post a Comment