TEKST ZADATKA
Skratiti razlomke (zadaci 188-189):
3 x 2 + 2 x − 8 12 x 2 − 7 x − 12 \frac{3x^2 + 2x - 8}{12x^2 - 7x - 12} 12 x 2 − 7 x − 12 3 x 2 + 2 x − 8 REŠENJE ZADATKA
Da bismo skratili razlomak, potrebno je da rastavimo kvadratne trinome u brojiocu i imeniocu na linearne činioce koristeći formulu a x 2 + b x + c = a ( x − x 1 ) ( x − x 2 ) . ax^2 + bx + c = a(x - x_1)(x - x_2) . a x 2 + b x + c = a ( x − x 1 ) ( x − x 2 ) .
Prvo rešavamo kvadratnu jednačinu u brojiocu 3 x 2 + 2 x − 8 = 0. 3x^2 + 2x - 8 = 0 . 3 x 2 + 2 x − 8 = 0. Računamo diskriminantu i njena rešenja primenom formule.
x 1 , 2 = − 2 ± 2 2 − 4 ⋅ 3 ⋅ ( − 8 ) 2 ⋅ 3 = − 2 ± 4 + 96 6 = − 2 ± 100 6 x_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-8)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 96}}{6} = \frac{-2 \pm \sqrt{100}}{6} x 1 , 2 = 2 ⋅ 3 − 2 ± 2 2 − 4 ⋅ 3 ⋅ ( − 8 ) = 6 − 2 ± 4 + 96 = 6 − 2 ± 100 Rešenja jednačine u brojiocu su:
x 1 = − 2 + 10 6 = 8 6 = 4 3 , x 2 = − 2 − 10 6 = − 12 6 = − 2 x_1 = \frac{-2 + 10}{6} = \frac{8}{6} = \frac{4}{3}, \quad x_2 = \frac{-2 - 10}{6} = \frac{-12}{6} = -2 x 1 = 6 − 2 + 10 = 6 8 = 3 4 , x 2 = 6 − 2 − 10 = 6 − 12 = − 2 Zapisujemo brojilac u faktorisanom obliku, ubacujući dobijena rešenja u formulu.
3 x 2 + 2 x − 8 = 3 ( x − 4 3 ) ( x − ( − 2 ) ) = ( 3 x − 3 ⋅ 4 3 ) ( x + 2 ) = ( 3 x − 4 ) ( x + 2 ) 3x^2 + 2x - 8 = 3\left(x - \frac{4}{3}\right)(x - (-2)) = \left(3x - 3 \cdot \frac{4}{3}\right)(x + 2) = (3x - 4)(x + 2) 3 x 2 + 2 x − 8 = 3 ( x − 3 4 ) ( x − ( − 2 )) = ( 3 x − 3 ⋅ 3 4 ) ( x + 2 ) = ( 3 x − 4 ) ( x + 2 ) Sada rešavamo kvadratnu jednačinu u imeniocu 12 x 2 − 7 x − 12 = 0. 12x^2 - 7x - 12 = 0 . 12 x 2 − 7 x − 12 = 0.
x 1 , 2 = − ( − 7 ) ± ( − 7 ) 2 − 4 ⋅ 12 ⋅ ( − 12 ) 2 ⋅ 12 = 7 ± 49 + 576 24 = 7 ± 625 24 x_{1,2} = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot (-12)}}{2 \cdot 12} = \frac{7 \pm \sqrt{49 + 576}}{24} = \frac{7 \pm \sqrt{625}}{24} x 1 , 2 = 2 ⋅ 12 − ( − 7 ) ± ( − 7 ) 2 − 4 ⋅ 12 ⋅ ( − 12 ) = 24 7 ± 49 + 576 = 24 7 ± 625 Rešenja jednačine u imeniocu su:
x 1 = 7 + 25 24 = 32 24 = 4 3 , x 2 = 7 − 25 24 = − 18 24 = − 3 4 x_1 = \frac{7 + 25}{24} = \frac{32}{24} = \frac{4}{3}, \quad x_2 = \frac{7 - 25}{24} = \frac{-18}{24} = -\frac{3}{4} x 1 = 24 7 + 25 = 24 32 = 3 4 , x 2 = 24 7 − 25 = 24 − 18 = − 4 3 Zapisujemo imenilac u faktorisanom obliku. Množilac 12 12 12 možemo rastaviti kao 3 ⋅ 4 3 \cdot 4 3 ⋅ 4 kako bismo eliminisali razlomke u zagradama.
12 x 2 − 7 x − 12 = 12 ( x − 4 3 ) ( x − ( − 3 4 ) ) = 3 ( x − 4 3 ) ⋅ 4 ( x + 3 4 ) = ( 3 x − 4 ) ( 4 x + 3 ) 12x^2 - 7x - 12 = 12\left(x - \frac{4}{3}\right)\left(x - \left(-\frac{3}{4}\right)\right) = 3\left(x - \frac{4}{3}\right) \cdot 4\left(x + \frac{3}{4}\right) = (3x - 4)(4x + 3) 12 x 2 − 7 x − 12 = 12 ( x − 3 4 ) ( x − ( − 4 3 ) ) = 3 ( x − 3 4 ) ⋅ 4 ( x + 4 3 ) = ( 3 x − 4 ) ( 4 x + 3 ) Zamenjujemo faktorisani brojilac i imenilac u početni razlomak.
3 x 2 + 2 x − 8 12 x 2 − 7 x − 12 = ( 3 x − 4 ) ( x + 2 ) ( 3 x − 4 ) ( 4 x + 3 ) \frac{3x^2 + 2x - 8}{12x^2 - 7x - 12} = \frac{(3x - 4)(x + 2)}{(3x - 4)(4x + 3)} 12 x 2 − 7 x − 12 3 x 2 + 2 x − 8 = ( 3 x − 4 ) ( 4 x + 3 ) ( 3 x − 4 ) ( x + 2 ) Skraćujemo razlomak sa zajedničkim činiocem 3 x − 4 3x - 4 3 x − 4 (uz uslov da je x ≠ 4 3 x \neq \frac{4}{3} x = 3 4 i x ≠ − 3 4 x \neq -\frac{3}{4} x = − 4 3 kako bi imenilac bio različit od nule) i dobijamo konačno rešenje.
x + 2 4 x + 3 \frac{x + 2}{4x + 3} 4 x + 3 x + 2