I.D.M MATEMATIKA FULL

Persamaan dan Fungsi Kuadrat

Author: Ilham D. Matalauni Genre:

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3. Persamaan kuadrat $x^{2} - 5x + 6 = 0$ mempunyai akar-akar $x_{1}$ dan $x_{2}$ . Persamaan kuadrat yang akar-akarnya $\left ( x_{1} - 3 \right )$ dan $\left ( x_{2} - 3 \right )$ adalah ...

a. $x^{2} - 2x = 0$

b. $x^{2} - 2x + 30 = 0$

c. $x^{2} + x = 0$

d. $x^{2} + x - 30 = 0$

e. $x^{2} + x + 30 = 0$

Jawaban :

=> Diketahui: $x_{1}$ dan $x_{2}$ akar-akar persamaan $x^{2} - 5x + 6 = 0,$ maka berlaku:

$x_{1} + x_{2} = \frac{-b}{a} = -\frac{(-5)}{1} = 5 ... (i)$

$x_{1} . x_{2} = \frac{c}{a} = \frac{(6)}{1} = 6 ... (ii)$

=> Dengan menggunakan persamaan (i) dan (ii), maka persamaan kuadrat yang akar-akarnya
$y_{1} = x_{1} - 3$ dan $y_{2} = x_{2} - 3,$ adalah:

$=> x^{2} - \left ( y_{1} + y_{2} \right )x + y_{1}.y_{2} = 0$

$=> x^{2} - \left \{ \left ( x_{1}-3 \right )+ (x_{2}-3) \right \}x + \left \{ \left ( x_{1}-3 \right ) . (x_{2}-3) \right \} = 0$

$=> x^{2} - \left \{ \left ( x_{1} + x_{2} \right )- 6 \right \}x + \left \{ x_{1} . x_{2} - 3\left ( x_{1} + x_{2}\right ) + 9\right \} = 0$

$=> x^{2} - \left \{ 5 - 6 \right \}x + \left \{ 6 - 3\left ( 5 \right ) + 9 \right \} = 0$

$=> x^{2} - \left \{ -1 \right \}x + \left \{ 6 - 15 + 9 \right \} = 0$

$=> x^{2} + x = 0$

Jawabannya adalah 'b'. Demikian soal no 3, Next no 4

Posted by Ilham D. Matalauni Posted on 0 comments

Logaritma

Author: Ilham D. Matalauni Genre:

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2. Jika $^2{\log }3 = a$    dan $^3{\log }5 = b,$ $^1^5{\log }20 = ...$

a.      $\frac{2}{a}$

b.    $\frac{2 + ab}{a(1+b)}$

c.     $\frac{a}{2}$

d.    $\frac{b + 1}{2ab + 1}$

e.    $\frac{a(1 + b)}{2 + ab}$

Jawaban :

Ingat sifat logaritma :

$^a{\log b } . ^b{\log c } = ^a{\log c }$

$^m{\log a x b } = ^m{\log a } + ^m{\log b }$

$^a{\log b } = \frac{{\log b }}{{\log a }} = \frac{1}{^b{\log a }}$

$^a{\log b } = \frac{1}{n}^a{\log b }$

=>

$^2{\log 3} = a ......... (i)$
$^3{\log 5} = b ......... (ii)$

=>

Dari pers. (i) dan (ii) diperoleh:

$^2{\log 3} . ^3{\log 5} = ^2{\log 5}$

$a . b = ^2{\log 5}$

$^2{\log 5} = a . b ......... (iii)$

$^1^5{\log 20} = ^1^5{\log (4 x 5)}$

$^1^5{\log 4} + ^1^5{\log 5}$

$\frac{1}{^4{\log 15}} + \frac{1}{^5{\log 15}}$

$\frac{1}{^2^{^2}{\log 15}} + \frac{1}{^5{\log 15}}$

$\frac{1}{\frac{1}{2}^2\log 15} + \frac{1}{^5{\log (3 x 5) }}$

$2 . \frac{1}{^2\log 3 + ^2\log 5} + \frac{1}{^5\log 3 + ^5\log 5} .......... (iv)$

=>

Subtitusi pers. (i), (ii), dan (iii) ke pers. (iv), sehingga diperoleh:

$^1^5\log 20 = 2 . \left ( \frac{1}{a + ab} \right ) + \frac{1}{\frac{1}{b} + 1}$

$= \frac{2}{a + ab} + \frac{b}{1 + b}$

$= \frac{2(1 + b) + b(a +ab)}{(a + ab)(1 + b)}$

$= \frac{2(1 + b) + ab(1 +b)}{(a + ab)(1 + b)}$

$= \frac{(1 + b) . (2 + ab)}{a(1 + b)(1 + b)}$

$= \frac{2 + ab}{a(1 + b)}$

Jawaban : option "b". Demikian soal no 2, Next no. 3