Operasi Aljabar pada Bentuk Akar dan Pembahasan Soal

a. Penjumlahan dan Pengurangan Bentuk Akar

Sebelum melakukan operasi penjumlahan dan pengurangan bentuk akar, kita harus memahami terlebih dahulu tentang akar senama dan akar sejenis.

Akar senama adalah akar-akar yang memiliki indeks sama. Sebagai contoh: $\sqrt[3]{2}$ , $\sqrt[3]{5}$ , $\sqrt[3]{7}$ , $\sqrt[3]{19}$ , $\sqrt[3]{x}$ , dan sebagainya.
Akar sejenis akar-akar yang memiliki indeks maupun radikan (bilangan pokok) sama. Sebagai contoh: $\sqrt[3]{2}$ , $8\sqrt[3]{2}$ , $\frac{1}{2} \sqrt[3]{2}$ , $x\sqrt[3]{2}$ , dan sebagainya.

Teorema:

Penjumlahan dan pengurangan bentuk akar dapat dilakukan juka akar-akarnya sejenis.

$a\sqrt[n]{c}+b \sqrt[n]{c}=(a+b) \sqrt[n]{c}$
$a\sqrt[n]{c}-b \sqrt[n]{c}=(a-b) \sqrt[n]{c}$

b. Perkalian dan Pembagian Bentuk Akar

Teorema:

Perkalian dan pembagian bentuk akar dapat dilakukan jika akar-akarnya senama.

$x\sqrt[n]{a} \times y \sqrt[n]{b}=xy \sqrt[n]{ab}$
$\left (a\sqrt[n]{b} \right)^n = a^nb$
$\sqrt{a} \times \sqrt{a} = ( \sqrt{a})^2 = a$ , dengan a ≥ 0
$x\sqrt[m]{a} \times y \sqrt[n]{b}$ = $x\sqrt[mn]{a^n} \times y \sqrt[mn]{b^m}$ = $xy\sqrt[mn]{a^nb^m}$
$\frac{x\sqrt[n]{a}}{y \sqrt[n]{b}}= \frac{x}{y} \sqrt[n]{\frac{a}{b}}$
$\frac{x\sqrt[m]{a}}{y \sqrt[n]{b}}$ = $\frac{x\sqrt[mn]{a^n}}{y \sqrt[mn]{b^m}}$ = $\frac{x}{y} \sqrt[mn]{\frac{a^n}{b^m}}$
$\frac{\sqrt[m]{a} \times \sqrt[m]{b}}{\sqrt[m]{c} \times \sqrt[m]{d}}= \sqrt[m]{\frac{ab}{cd}}$

c. Akar dari Suku Dua yang Kedua Sukunya Merupakan Bentuk Akar

Teorema:

Jika a > 0, b > 0, c > 0, dan c bilangan rasional positif, maka

$\sqrt{(a + b)+2 \sqrt{ab}} = \sqrt{a} + \sqrt{b}$
$\sqrt{(a + b)-2 \sqrt{ab}} = \sqrt{a} - \sqrt{b}$ , dengan a > b
$\sqrt{a+b \sqrt{c}}= \sqrt{\frac{a+p}{2}}+ \sqrt{\frac{a-p}{2}}$ , dengan $p=\sqrt{a^2 -(b \sqrt{c})^2}$
$\sqrt{a-b \sqrt{c}}= \sqrt{\frac{a+p}{2}}- \sqrt{\frac{a-p}{2}}$ , dengan $p=\sqrt{a^2 -(b \sqrt{c})^2}$

Contoh Soal 1

Sederhanakanlah.

$8\sqrt{2} + 6 \sqrt{2}$

$15\sqrt{3} - 16 \sqrt{3}$

$3 \sqrt{5}+ \sqrt{20} - 2 \sqrt{125}$

$3\sqrt{8} - \sqrt{12}+ 18 \sqrt{\frac{1}{2}}$

Jawab:

$8\sqrt{2} + 6 \sqrt{2}$

$(8 + 6)\sqrt{2}$ = $14\sqrt{2}$

$15\sqrt{3} - 16 \sqrt{3}$

$(15-16)\sqrt{3}$ = $-\sqrt{3}$

$3 \sqrt{5}+ \sqrt{20} - 2 \sqrt{125}$

$3 \sqrt{5}+ \sqrt{4 \times 5} - 2 \sqrt{25 \times 5}$

$3 \sqrt{5}+ 2\sqrt{5} - 2 \times 5\sqrt{5}$

= $(3 + 2 - 10) \sqrt{5}$ = $-5 \sqrt{5}$

$3\sqrt{8} - \sqrt{12}+ 18 \sqrt{\frac{1}{2}}$

$3\sqrt{4 \times 2} - \sqrt{4 \times 3}+ 18 \times \frac{1}{2}\sqrt{2}$

= $6\sqrt{2} - 2\sqrt{3}+ 9 \sqrt{2}$

Baca Juga

= $(6 + 9) \sqrt{2} - 2 \sqrt{3}= 15 \sqrt{2} - 2 \sqrt{3}$

Contoh Soal 2

Tentukan hasil perkalian berikut dalam bentuk akar paling sederhana.

$\sqrt{7} \times \sqrt{3}$

$2\sqrt{5} \times \sqrt{15}$

$2 \sqrt[3]{12} \times 4 \sqrt[3]{6}$

$5\sqrt[3]{2} \times 4 \sqrt{3}$

Jawab:

$\sqrt{7} \times \sqrt{3} = \sqrt{7 \times 3} = \sqrt{21}$

$2\sqrt{5 \times 15}$

= $2\sqrt{5^2 \times 3} = 2 \times 5 \sqrt{3}= 10 \sqrt{3}$

$2 \sqrt[3]{12} \times 4 \sqrt[3]{6}= 8 \sqrt[3]{12 \times 6}$

$8 \sqrt[]{2^3 \times 3^2}= 8 \times 2\sqrt[3]{3^2}=16 \sqrt[3]{9}$

$5\sqrt[3]{2} \times 4 \sqrt{3}=20 \sqrt[6]{2^2} \times \sqrt[6]{3^3}$

$20 \sqrt[6]{2^2 \times 3^3}=20 \sqrt[6]{108}$

Contoh Soal 3

Tentukan hasil perkalian berikut dalam bentuk akar paling sederhana.

$\sqrt{3}(\sqrt{6} +2 \sqrt{3})$

$(2\sqrt{2} -5) 4\sqrt{2}$

$(\sqrt{5} + 2 \sqrt{3})(\sqrt{5} - 2 \sqrt{3})$

$(5\sqrt{3} - \sqrt{2})(5\sqrt{3} + 2 \sqrt{2})$

$(6\sqrt{6} + 3)(2\sqrt{3} - \sqrt{2})$

$(5-6 \sqrt{5})^2$

$(\sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3} - \sqrt{5})$

$(\sqrt{2}+ \sqrt{3} + \sqrt{5})^2$

$(\sqrt[3]{4}-2)^3$

$(2\sqrt[3]{6}- \sqrt[3]{3})(4 \sqrt[3]{36}+2 \sqrt{28} + \sqrt[3]{9})$

Jawab:

$\sqrt{3}(\sqrt{6} +2 \sqrt{3})$

$\sqrt{3} \times \sqrt{6} + \sqrt{3} \times 2 \sqrt{3})$

= $\sqrt{18} + 2 \times 3 = 3 \sqrt{2} + 6$

$(2\sqrt{2} -5) 4\sqrt{2}$

$2\sqrt{2} \times 4 \sqrt{2} -5 \times 4\sqrt{2}$

= $8 \times 2 - 20 \sqrt{2} = 16 - 20 \sqrt{2}$

$(\sqrt{5} + 2 \sqrt{3})(\sqrt{5} - 2 \sqrt{3})$

$( \sqrt{5})^2 - (2 \sqrt{3})^2$

= $5 - 2^2 \times 3$ = 5 - 12 = -7

$(5\sqrt{3} - \sqrt{2})(5\sqrt{3} + 2 \sqrt{2})$

$(5\sqrt{3})^2 + (- \sqrt{2}+2 \sqrt{2})5 \sqrt{2} - \sqrt{2} \times 2 \sqrt{2}$

= $5^2 \times 3 + \sqrt{2} \times 5 \sqrt{3} - 2 \times 2$

= $71 + 5 \sqrt{6}$

$(6\sqrt{6} + 3)(2\sqrt{3} - \sqrt{2})$

$6\sqrt{6} \times 2 \sqrt{3} - 6 \sqrt{6} \times - \sqrt{2}$ + $3 \times 2 \sqrt{3} - 3 \times \sqrt{2}$

$12 \sqrt{18}- 6 \sqrt{12} + 6 \sqrt{3} - 3 \sqrt{2}$

$36 \sqrt{2}-12 \sqrt{3} + 6 \sqrt{3} - 3 \sqrt{2}$

$33 \sqrt{2}- 6 \sqrt{3}$

$(5-6 \sqrt{5})^2$

$5^2 -2 \times 5 \times 6 \sqrt{5} + (6 \sqrt{5})^2$

$25 - 60 \sqrt{5} + 180$

$205 - 60 \sqrt{5}$

$(\sqrt{2} + \sqrt{3} + \sqrt{5})(\sqrt{2} + \sqrt{3} - \sqrt{5})$

$(\sqrt{2}+ \sqrt{3})^2 - (\sqrt{5})^2$

$2 + 2 \sqrt{6} + 3 - 5 = 2 \sqrt{6}$

$(\sqrt{2}+ \sqrt{3} + \sqrt{5})^2$

$(\sqrt{2})^2 + (\sqrt{3})^2 + (\sqrt{5})^2 + 2 \sqrt{2} \times \sqrt{3}$ +

$2 \sqrt{2} \times \sqrt{3} \times \sqrt{5}$

$2 + 3 + 5 + 2 \sqrt{6} + 2 \sqrt{10} + 2 \sqrt{15}$

$10 + 2 \sqrt{6} + 2 \sqrt{10} + 2 \sqrt{15}$

$(\sqrt[3]{4}-2)^3$

$(\sqrt[3]{4})^3 - 3 (\sqrt[3]{4})^2 \times 2$ +

$3 \sqrt[3]{4}(2)^2 - 2^3$

$4 - 6 \sqrt[3]{16} + 12 \sqrt[3]{4} - 8$

$-4-12 \sqrt[3]{2} + 12 \sqrt[3]{4}$

$(2\sqrt[3]{6}- \sqrt[3]{3})(4 \sqrt[3]{36}+2 \sqrt{28} + \sqrt[3]{9})$

$(2 \sqrt[3]{6})^3 - (\sqrt[3]{3})^3$

$2^3 \times 6 - 3 = 45$

Contoh Soal 4

Sederhanakanlah!

$\sqrt{8 + 2 \sqrt{15}}$

$\sqrt{12 - 2 \sqrt{35}}$

$\sqrt{9 + \sqrt{56}}$

$\sqrt{14 - 6 \sqrt{5}}$

$\sqrt{7 + 4\sqrt{3}}$

$\sqrt{18 - \sqrt{260}}$

$\sqrt{3 + \sqrt{5}}$

$\sqrt{8 - 3\sqrt{7}}$

Jawab:

$\sqrt{8 + 2 \sqrt{15}} = \sqrt{5} + \sqrt{3}$

$\sqrt{12 - 2 \sqrt{35}} = \sqrt{7} - \sqrt{5}$

$\sqrt{9 + \sqrt{56}} = \sqrt{9 + 2\sqrt{14}} = \sqrt{7} + \sqrt{2}$

$\sqrt{14 - 6 \sqrt{5}}= \sqrt{14-2 \sqrt{45}}$

$\sqrt{9} - \sqrt{5}= 3 - \sqrt{5}$

$\sqrt{7 + 4\sqrt{3}}= \sqrt{7 + 2 \sqrt12}$

$\sqrt{4} + \sqrt{3} = 2 + \sqrt{3}$

$\sqrt{18 - \sqrt{260}} = \sqrt{18 - 2 \sqrt{65}}$

$\sqrt{13} - \sqrt{5}$

$p = \sqrt{3^2 - (\sqrt{5})^2}= \sqrt{4} = 2$

$\sqrt{3 + \sqrt{5}} = \sqrt{\frac{3+p}{2}} + \sqrt{\frac{3-p}{2}}$

= $\sqrt{\frac{3+2}{2}+ \frac{3-2}{2}}$

= $\sqrt{\frac{5}{2}} + \sqrt{\frac{1}{2}}$

= $\frac{1}{2} \sqrt{10} + \frac{1}{2} \sqrt{2}$

= $\frac{1}{2}(\sqrt{10} + \sqrt{2})$

h. $p = \sqrt{8^2 - (3 \sqrt{7})^2} = 1$

$\sqrt{8 - 3\sqrt{7}} = \sqrt{\frac{8+p}{2}}- \sqrt{\frac{8-p}{2}}$

= $\sqrt{\frac{8+1}{2}} - \sqrt{\frac{8-1}{2}}$

= $\sqrt{\frac{9}{2}}- \sqrt{\frac{7}{2}}$

= $\frac{3}{2} \sqrt{2} - \frac{1}{2} \sqrt{14}$

= $\frac{1}{2} (3 \sqrt{2} - \sqrt{14})$

Location: