Podaj pierwiastki trójmianu kwadratowego i zapisz go w postaci ogólnej - Zadanie 1: MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

Rozwiązanie

$a)$

$x + 10 = 0 l u b x - 20 = 0$

$x = - 10 l u b x = 20$

$y = (x + 10) (x - 20) = x (x - 20) + 10 (x - 20) =$

$= x^{2} - 20 x + 10 x - 200 = x^{2} - 10 x - 200$

$b)$

$x - \frac{1}{2} = 0 l u b x - \frac{1}{3} = 0$

$x = \frac{1}{2} l u b x = \frac{1}{3}$

$y = 6 (x - \frac{1}{2}) (x - \frac{1}{3}) =$ $(6 x - 3) (x - \frac{1}{3}) =$

$= 6 x (x - \frac{1}{3}) - 3 (x - \frac{1}{3}) =$

$= 6 x^{2} - 2 x - 3 x + 1 =$ $6 x^{2} - 5 x + 1$

$c)$

$x + 3 = 0 l u b 3 x - 4 = 0$

$x = - 3 l u b 3 x = 4$

$x = \frac{4}{3} = 1 \frac{1}{3}$

$y = 2 (x + 3) (3 x - 4) =$ $(2 x + 6) (3 x - 4) =$

$= 2 x (3 x - 4) + 6 (3 x - 4) =$

$= 6 x^{2} - 8 x + 18 x - 24 =$

$= 6 x^{2} + 10 x - 24$

$d)$

$x - \frac{2}{5} = 0 l u b x + \frac{2}{5} = 0$

$x = \frac{2}{5} l u b x = - \frac{2}{5}$

$y = 5 (x - \frac{2}{5}) (x + \frac{2}{5}) =$ $(5 x - 2) (x + \frac{2}{5}) =$

$= 5 x (x + \frac{2}{5}) - 2 (x + \frac{2}{5}) =$

$= 5 x^{2} + 2 x - 2 x - \frac{4}{5} =$ $5 x^{2} - \frac{4}{5}$

$e)$

$2 x - 4 = 0 l u b 2 x - 1 = 0$

$2 x = 4 l u b 2 x = 1$

$x = 2 l u b x = \frac{1}{2}$

$y = (2 x - 4) (2 x - 1) = 2 x (2 x - 1) - 4 (2 x - 1) =$

$= 4 x^{2} - 2 x - 8 x + 4 =$ $4 x^{2} - 10 x + 4$

$f)$

$2 x + 1 = 0 l u b 3 x + 1 = 0$

$2 x = - 1 l u b 3 x = - 1$

$x = - \frac{1}{2} l u b x = - \frac{1}{3}$

$y = (2 x + 1) (3 x + 1) = 2 x (3 x + 1) + 1 (3 x + 1) =$

$= 6 x^{2} + 2 x + 3 x + 1 =$ $6 x^{2} + 5 x + 1$

$g)$

$5 x - 2 = 0 l u b 4 x - 3 = 0$

$5 x = 2 l u b 4 x = 3$

$x = \frac{2}{5} l u b x = \frac{3}{4}$

$y = (5 x - 2) (4 x - 3) =$ $5 x (4 x - 3) - 2 (4 x - 3) =$

$= 20 x^{2} - 15 x - 8 x + 6 =$ $20 x^{2} - 23 x + 6$

$h)$

$2 - 3 x = 0 l u b x + 4 = 0$

$2 = 3 x l u b x = - 4$

$x = \frac{2}{3}$

$y = (2 - 3 x) (x + 4) = 2 (x + 4) - 3 x (x + 4) =$

$= 2 x + 8 - 3 x^{2} - 12 x =$ $- 3 x^{2} - 10 x + 8$

$i)$

$x = 0 l u b x + 3 = 0$

$x = - 3$

$y = - x (x + 3) = - x^{2} - 3 x$

$j)$

$\frac{1}{2} x = 0 l u b 2 x - 1 = 0$

$x = 0 l u b 2 x = 1$

$x = \frac{1}{2}$

$y = \frac{1}{2} x (2 x - 1) = x^{2} - \frac{1}{2} x$

$k)$

$7 - 3 x = 0 l u b x = 0$

$7 = 3 x$

$x = \frac{7}{3} = 2 \frac{1}{3}$

$y = (7 - 3 x) x = 7 x - 3 x^{2} = - 3 x^{2} + 7 x$

$l)$

$2 x + 3 = 0$

$2 x = - 3$

$x = - \frac{3}{2} = - 1 \frac{1}{2}$

$y = (2 x + 3)^{2} = (2 x)^{2} + 2 \cdot 2 x \cdot 3 + 3^{2} =$

$= 4 x^{2} + 12 x + 9$

Agnieszka Nowak

Nauczycielka matematyki

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