Klasa
II liceum
Przedmiot
Matematyka
Wybierz książkę
Matematyka 2. Poziom rozszerzony. Po gimnazjum, Zbiór zadań
  • 2.34

    Zadanie

  • 2.35

    Zadanie

  • 2.36

    Zadanie

  • 2.37

    Zadanie

  • 2.38

    Zadanie

  • 2.39

    Zadanie

Przydatne będą wzory: 

 

Δ=b24ac\Delta={b}^{{2}}-{4}{a}{c} 

W=(p, q),    p=b2a,    q=Δ4a{W}={\left({p},\ {q}\right)},\ \ \ \ {p}=\frac{{-{b}}}{{{2}{a}}},\ \ \ \ {q}=\frac{{-\Delta}}{{{4}{a}}}   

 

 

 

a){a}{)} 

 

 Δ=12      b24ac=12      824(1)c=12      64+4c=12  64      4c=52  4      c=13\Delta={12}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}={12}\ \ \ \Rightarrow\ \ \ {8}^{{2}}-{4}\cdot{\left(-{1}\right)}\cdot{c}={12}\ \ \ \Rightarrow\ \ \ {64}+{4}{c}={12}\ \ {\mid}-{64}\ \ \ \Rightarrow\ \ \ {4}{c}=-{52}\ \ {\left|{4}\ \ \ \Rightarrow\ \ \ {c}=-{13}\right.}

f(x)=x2+8x13\underline{{\underline{{{f{{\left({x}\right)}}}=-{x}^{{2}}+{8}{x}-{13}}}}}        

 

 

 

b){b}{)} 

1=Δ4a      1=84a  4a      4a=8  4      a=2{1}=\frac{{-\Delta}}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ {1}=\frac{{8}}{{{4}{a}}}\ \ {\mid}\cdot{4}{a}\ \ \ \Rightarrow\ \ \ {4}{a}={8}\ \ {\left|{4}\ \ \ \Rightarrow\ \ \ {a}={2}\right.} 

12=b2a      12=b22   (4)      b=2\frac{{1}}{{2}}=\frac{{-{b}}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ \frac{{1}}{{2}}=\frac{{-{b}}}{{{2}\cdot{2}}}\ \ \ {\mid}\cdot{\left(-{4}\right)}\ \ \ \Rightarrow\ \ \ {b}=-{2} 

Δ=8      b24ac=8      (2)242c=8  4      8c=12  (8)      c=128=32=112\Delta=-{8}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}=-{8}\ \ \ \Rightarrow\ \ \ {\left(-{2}\right)}^{{2}}-{4}\cdot{2}\cdot{c}=-{8}\ \ {\mid}-{4}\ \ \ \Rightarrow\ \ \ -{8}{c}=-{12}\ \ {\left|{\left(-{8}\right)}\ \ \ \Rightarrow\ \ \ {c}=\frac{{12}}{{8}}=\frac{{3}}{{2}}={1}\frac{{1}}{{2}}\right.} 

f(x)=2x22x+112\underline{{\underline{{{f{{\left({x}\right)}}}={2}{x}^{{2}}-{2}{x}+{1}\frac{{1}}{{2}}}}}} 

 

 

 

c){c}{)} 

5=Δ4a      5=Δ4a      5=1204a  4a      20a=120  20      a=6-{5}=-\frac{\Delta}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ {5}=\frac{\Delta}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ {5}=\frac{{120}}{{{4}{a}}}\ \ {\mid}\cdot{4}{a}\ \ \ \Rightarrow\ \ \ {20}{a}={120}\ \ {\left|{20}\ \ \ \Rightarrow\ \ \ {a}={6}\right.} 

0=b2a      0=b26  (12)      b=0{0}=-\frac{{b}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ {0}=-\frac{{b}}{{{2}\cdot{6}}}\ \ {\mid}\cdot{\left(-{12}\right)}\ \ \ \Rightarrow\ \ \ {b}={0} 

Δ=120      b24ac=120      0246c=120   (24)      c=5\Delta={120}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}={120}\ \ \ \Rightarrow\ \ \ {0}^{{2}}-{4}\cdot{6}\cdot{c}={120}\ \ \ {\left|{\left(-{24}\right)}\ \ \ \Rightarrow\ \ \ {c}=-{5}\right.} 

f(x)=6x25\underline{{\underline{{{f{{\left({x}\right)}}}={6}{x}^{{2}}-{5}}}}} 

 

 

 

d){d}{)} 

3=Δ4a      3=364a  4a      12a=36  (12)      a=3-{3}=-\frac{\Delta}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ -{3}=-\frac{{36}}{{{4}{a}}}\ \ {\mid}\cdot{4}{a}\ \ \ \Rightarrow\ \ \ -{12}{a}=-{36}\ \ {\left|{\left(-{12}\right)}\ \ \ \Rightarrow\ \ \ {a}={3}\right.} 

1=b2a      1=b2a      1=b23  6      b=6-{1}=-\frac{{b}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ {1}=\frac{{b}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ {1}=\frac{{b}}{{{2}\cdot{3}}}\ \ {\mid}\cdot{6}\ \ \ \Rightarrow\ \ \ {b}={6} 

Δ=36      b24ac=36      6243c=36  36      12c=0      c=0\Delta={36}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}={36}\ \ \ \Rightarrow\ \ \ {6}^{{2}}-{4}\cdot{3}\cdot{c}={36}\ \ {\mid}-{36}\ \ \ \Rightarrow\ \ \ -{12}{c}={0}\ \ \ \Rightarrow\ \ \ {c}={0} 

f(x)=3x2+6x\underline{{\underline{{{f{{\left({x}\right)}}}={3}{x}^{{2}}+{6}{x}}}}} 

 

 

 

e){e}{)} 

614=Δ4a      254=254a  4a      25a=25      a=1-{6}\frac{{1}}{{4}}=-\frac{\Delta}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ -\frac{{25}}{{4}}=-\frac{{25}}{{{4}{a}}}\ \ {\mid}\cdot{4}{a}\ \ \ \Rightarrow\ \ \ -{25}{a}=-{25}\ \ \ \Rightarrow\ \ \ {a}={1} 

32=b2a      32=b2   (2)      b=3\frac{{3}}{{2}}=-\frac{{b}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ \frac{{3}}{{2}}=-\frac{{b}}{{2}}\ \ \ {\mid}\cdot{\left(-{2}\right)}\ \ \ \Rightarrow\ \ \ {b}=-{3} 

Δ=25      b24ac=25      (3)241c=25   9      4c=16  (4)      c=4\Delta={25}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}={25}\ \ \ \Rightarrow\ \ \ {\left(-{3}\right)}^{{2}}-{4}\cdot{1}\cdot{c}={25}\ \ \ {\mid}-{9}\ \ \ \Rightarrow\ \ \ -{4}{c}={16}\ \ {\left|{\left(-{4}\right)}\ \ \ \Rightarrow\ \ \ {c}=-{4}\right.} 

f(x)=x23x4\underline{{\underline{{{f{{\left({x}\right)}}}={x}^{{2}}-{3}{x}-{4}}}}} 

 

 

 

f){f}{)} 

120=Δ4a      120=14a      4a=20  4      a=5\frac{{1}}{{20}}=-\frac{\Delta}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ \frac{{1}}{{20}}=\frac{{1}}{{{4}{a}}}\ \ \ \Rightarrow\ \ \ {4}{a}={20}\ \ {\left|{4}\ \ \ \Rightarrow\ \ \ {a}={5}\right.} 

310=b2a      310=b25  (10)      b=3\frac{{3}}{{10}}=-\frac{{b}}{{{2}{a}}}\ \ \ \Rightarrow\ \ \ \frac{{3}}{{10}}=-\frac{{b}}{{{2}\cdot{5}}}\ \ {\mid}\cdot{\left(-{10}\right)}\ \ \ \Rightarrow\ \ \ {b}=-{3} 

Δ=1      b24ac=1      (3)245c=1  9   20c=10   (20)      c=1020=12\Delta=-{1}\ \ \ \Rightarrow\ \ \ {b}^{{2}}-{4}{a}{c}=-{1}\ \ \ \Rightarrow\ \ \ {\left(-{3}\right)}^{{2}}-{4}\cdot{5}\cdot{c}=-{1}\ \ {\mid}-{9}\ \ \ -{20}{c}=-{10}\ \ \ {\left|{\left(-{20}\right)}\ \ \ \Rightarrow\ \ \ {c}=\frac{{10}}{{20}}=\frac{{1}}{{2}}\right.} 

f(x)=5x23x+12\underline{{\underline{{{f{{\left({x}\right)}}}={5}{x}^{{2}}-{3}{x}+\frac{{1}}{{2}}}}}} 

Komentarze

Avatar komentatora
Piotrek17 grudnia 2018
Dziena 👍
0