Operasikan bentuk akar ini sampai diperoleh bentuk paling sederhana
[tex]\displaystyle \sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}-\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}[/tex]
[tex]\displaystyle \sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}-\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}[/tex]
Verified answer
Jawab:
[tex]=\bf\sqrt{2}[/tex]
Penjelasan dengan langkah-langkah:
Dari yg paling dalam dulu
[tex]\sqrt{13\pm4\sqrt{3}}[/tex]
Gunakan rumus
[tex]\sqrt{p+q\pm\sqrt{4pq}}=\sqrt{p}\pm\sqrt{q}\:\:\:\:\{p > q\}\\\sqrt{~~13~~\pm~4\sqrt{3}}=\sqrt{p}\pm\sqrt{q}[/tex]
Oleh karena itu
[tex]\displaystyle p+q=13,\:\:\:\:\sqrt{4pq}=4\sqrt{3}\\~~~~~~~~~~~~~~~~~~~\sqrt{4pq}=\sqrt{16}\sqrt{3}\\~~~~~~~~~~~~~~~~~~~\sqrt{4pq}=\sqrt{48}\\\\~~~~~~~~~~~~~~~~~~~~~~~~pq=\frac{48}{4}\\\\~~~~~~~~~~~~~~~~~~~~~~~~pq=12[/tex]
12 + 1 = 13 12(1) = 12 maka
[tex]\displaystyle\sqrt{~~13~~\pm~4\sqrt{3}}=\sqrt{12}\pm\sqrt{1}\\~~~~~~~~~~~~~~~~~~~~~=\sqrt{4}\sqrt{3}\pm1\\~~~~~~~~~~~~~~~~~~~~~=~~~2\sqrt{3}\pm1[/tex] maka
[tex]\displaystyle\sqrt{1+\sqrt{3+\bf\sqrt{13+4\sqrt{3}}} }-\rm\sqrt{1-\sqrt{3-\bf\sqrt{13-4\sqrt{3}}}}\\\\=\rm\sqrt{1+\sqrt{3+2\sqrt{3}+1} }-\sqrt{1-\sqrt{3-(2\sqrt{3}-1)} }\\\\=\sqrt{1+\sqrt{3+2\sqrt{3}+1} }-\sqrt{1-\sqrt{3-2\sqrt{3}+1} }\\\\=\sqrt{1+\sqrt{4+2\sqrt{3}} }-\sqrt{1-\sqrt{4-2\sqrt{3}} }[/tex]
Lalu selesaikan bentuk akar selanjutnya
[tex]\sqrt{4\pm2\sqrt{3}}[/tex]
Gunakan rumus
[tex]\sqrt{p+q\pm2\sqrt{pq}}=\sqrt{p}\pm\sqrt{q}\:\:\:\:\{p > q\}\\\sqrt{~~4~~~\pm2\sqrt{3}~~}=\sqrt{p}\pm\sqrt{q}[/tex]
Oleh karena itu
[tex]\displaystyle p+q=4,\:\:\:\:\:\:\:\:\:\:\:\:pq=3[/tex]
3 + 1 = 4, 3(1) = 3 maka
[tex]\displaystyle\sqrt{~~4~~~\pm2\sqrt{3}~~}=\sqrt{3}\pm\sqrt{1}\\~~~~~~~~~~~~~~~~~~~~~~=\sqrt{3}\pm1[/tex] maka
[tex]\displaystyle\sqrt{1+\bf\sqrt{4+2\sqrt{3}} }-\rm\sqrt{1-\bf\sqrt{4-2\sqrt{3}} }\rm\\\\=\sqrt{1+\sqrt{3}+1}-\sqrt{1-(\sqrt{3}-1)}\\\\=\sqrt{2+\sqrt{3}}-\sqrt{1-\sqrt{3}+1}\\\\=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}[/tex]
Selesaikan bentuk akar utk yg terakhir kalinya
[tex]\sqrt{2\pm\sqrt{3}}[/tex]
Gunakan rumus
[tex]\sqrt{p+q\pm\sqrt{4pq}}=\sqrt{p}\pm\sqrt{q}\:\:\:\:\{p > q\}\\\sqrt{~~2~~\pm~~~\sqrt{3}~}=\sqrt{p}\pm\sqrt{q}[/tex]
Oleh karena itu
[tex]\displaystyle p+q=2,\:\:\:\:\:\:\:\:\:\:\:\:\:\:4pq=3\\p+q=2,\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:pq=\frac{3}{4}[/tex]
Kali ini, teori Vieta digunakan. Jika
p dan q akar-akar pers. kuadrat dgn A = 1
[tex]\displaystyle x^2-(p+q)x\:\:\:\:\:\:\:\:=-pq\\\\x^2-~2x~~~~~~~~~~~~=-\frac{3}{4}\\\\x^2-~2x+\left(\frac{2}{2}\right)^2=-\frac{3}{4}+\left(\frac{2}{2}\right)^2\\\\x^2-~2x(1)+1^2\:\:=-\frac{3}{4}+1^2[/tex]
[tex]\displaystyle\because m^2-2mn+n^2=(m-n)^2\therefore\\\\~~~~~~~~~~~~~~(x-1)^2=-\frac{3}{4}+1\\\\~~~~~~~~~~~~~~(x-1)^2=-\frac{3}{4}+\frac{4}{4}\\\\~~~~~~~~~~~~~~(x-1)^2=\frac{1}{4}\\\\~~~~~~~~~~~~~~|\:x-1\:|\:=\sqrt{\frac{1}{4}}\\\\~~~~~~~~~~~~~~|\:x-1\:|\:=\frac{\sqrt1}{\sqrt4}\\\\~~~~~~~~~~~~~~|\:x-1\:|\:=\frac{1}{2}[/tex]
[tex]\because |m|=n\:\:\rightarrow\:\:m=\{n, -n\}\therefore[/tex]
p > q, maka
[tex]\displaystyle p=\frac{1}{2}+1,\:\:\:\:\:\:\:\:\:q=-\frac{1}{2}+1\\\\p=\frac{1}{2}+\frac{2}{2},\:\:\:\:\:\:\:\:\:q=-\frac{1}{2}+\frac{2}{2}\\\\p=\frac{3}{2},\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:q=\frac{1}{2}[/tex]
Oleh karena itu
[tex]\displaystyle\sqrt{~~2~~\pm~~~\sqrt{3}~}=~~~~~\:\sqrt{\frac{3}{2}}\pm\sqrt{\frac{1}{2}}\\\\~~~~~~~~~~~~~~~~~~~~~=~~~~~~\frac{\sqrt3}{\sqrt2}\pm\frac{1}{\sqrt2}\\\\~~~~~~~~~~~~~~~~~~~~~=\frac{\sqrt3(\sqrt2)}{(\sqrt2)^2}\pm\frac{\sqrt2}{(\sqrt2)^2}\\\\~~~~~~~~~~~~~~~~~~~~~=~~~~~~\:\frac{\sqrt6}{2}\pm\frac{\sqrt2}{2}\\\\~~~~~~~~~~~~~~~~~~~~~=~~~~~~~\frac{\sqrt6\pm\sqrt2}{2}[/tex] maka
[tex]\displaystyle\sqrt{2+\sqrt{3}}\:\:\:\:\:-\sqrt{2-\sqrt{3}}\\\\=\frac{\sqrt6+\sqrt2}{2}-\frac{\sqrt6-\sqrt2}{2}\\\\=\frac{\sqrt6+\sqrt2-(\sqrt6-\sqrt2)}{2}\\\\=\:\frac{\sqrt6+\sqrt2-\sqrt6+\sqrt2}{2}\\\\=~~~~~~~~~~~\:\frac{2\sqrt{2}}{2}\\\\=~~~~~~~~~~~~\:\bf\sqrt{2}[/tex]
(xcvi)