4 ponovimo! za realni broj x njegova je apsolutna vrijednost (modul) broj x koji odre đujemo na...

1

Zadatak 181 (Manuel, srednja škola)

Koliko iznosi zbroj rješenja jednadžbe ( ) ( ) ( )3 2

2 5 7 5 7 5 2 0?x x x⋅ + − ⋅ + + ⋅ + − =

33 31 25 23. . . .

2 2 2 2A B C D− − − −

Rješenje 181

Ponovimo!

( ) ( ) ( )33 3 2 2 3

,2

3 3 .2 3

a b a b a a b b a b a a b a b b− = − ⋅ + ⋅ + + = + ⋅ ⋅ + ⋅ ⋅ +

( )2 2

, , .2

1,2

n a c a d b c a c a d b ca b a a b b n

b d b d b d b d

⋅ + ⋅ ⋅ − ⋅+ = + ⋅ ⋅ + = + = − =

⋅ ⋅

Zakon distribucije množenja prema zbrajanju.

( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Skratiti razlomak znači brojnik i nazivnik tog razlomka podijeliti istim brojem različitim od nule i

jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Da bi umnožak bio jednak nuli, dovoljno je da jedan faktor bude jednak nuli.

0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =

Neka je

( )3 2

f x a x b x c x d= ⋅ + ⋅ + ⋅ +

polinom trećeg stupnja, a x1, x2, x3 njegove nultočke. Tada vrijede Vièteove formule:

1 2 3 1 2 1 3, ,

2.

3 1 2 3

b c dx x x x x x x x x x x x

a a a+ + = − ⋅ + ⋅ + ⋅ = ⋅ ⋅ = −

1.inačica

( ) ( ) ( )zamjena3 2 3 2

2 5 7 5 7 5 2 25

0 7 7 2 0x x x tt

tx

t+ =

⋅ + − ⋅ + + ⋅ + − = ⇒ ⇒ ⋅ − ⋅ + ⋅ − = ⇒

( ) ( )3 2 3

2 2 7 7 0 2 1 7 1 0t t t t t t⇒ ⋅ − − ⋅ + ⋅ = ⇒ ⋅ − − ⋅ ⋅ − = ⇒

( ) ( ) ( ) ( ) ( )( )2 22 1 1 7 1 0 1 2 1 7 0t t t t t t t t t⇒ ⋅ − ⋅ + + − ⋅ ⋅ − = ⇒ − ⋅ ⋅ + + − ⋅ = ⇒

( ) ( ) ( ) ( )1 02 2

1 2 2 2 7 0 1 2 5 2 0 .2

2 5 2 0

tt t t t t t t

t t

− =⇒ − ⋅ ⋅ + ⋅ + − ⋅ = ⇒ − ⋅ ⋅ − ⋅ + = ⇒

⋅ − ⋅ + =

• Riješimo prvu linearnu jednadžbu.

1 0 1.1

t t− = ⇒ =

• Riješimo drugu kvadratnu jednadžbu.

2 , 5 , 22

2 2 5 2 022 5 2 0

42 , 5 , 2

2,3 2

a b c

t tt t

b b a ca b c t

a

= = − =

⋅ − ⋅ + =⋅ − ⋅ + = ⇒ ⇒ ⇒

− ± − ⋅ ⋅= = − = =

⋅

( ) ( )2

5 5 4 2 2 5 25 16 5 9

2,3 2,3 2,32 2 4 4t t t

− − ± − − ⋅ ⋅ ± − ±⇒ = ⇒ = ⇒ = ⇒

⋅

2

5 3 82

22 2 25 3 4 4.12,3 5 3 24

3 23 3 34 4

8

4

2

4

tt t t

tt

t t t

+== = =

±⇒ = ⇒ ⇒ ⇒ ⇒

− == = =

Vraćamo se zamjeni.

1 45 1 1 51 1zamjena

2 5 2 2 5 32 2

1 11 1 55 53 2 22

5

2 1

t xx x

t x x x

x xt x

x t

= = −+ = = −

= ⇒ ⇒ + = ⇒ = − ⇒ = − ⇒

+ = = −= =

+ =

−

4 41 1

3 3 .2 2

1 10 9

32 2

x x

x x

x x

= − = −

⇒ = − ⇒ = −

−= = −

Zbroj rješenja jednadžbe iznosi:

( )9 9 4 3 9

4 3 4 31 2 3 1 2 3 1 2 32 2 1 1 2

x x x x x x x x x+ + = − + − + − ⇒ + + = − − − ⇒ + + = − − − ⇒

8 6 9 23.

1 2 3 1 2 32 2x x x x x x

− − −⇒ + + = ⇒ + + = −

Odgovor je pod D.

2.inačica

Preoblikujemo zadanu jednadžbu i uporabimo Vièteovu formulu.

( ) ( ) ( )3 2

2 5 7 5 7 5 2 0x x x⋅ + − ⋅ + + ⋅ + − = ⇒

( ) ( ) ( )3 2 2 3 2 2

2 3 5 3 5 5 7 2 5 5 7 5 2 0x x x x x x⇒ ⋅ + ⋅ ⋅ + ⋅ ⋅ + − ⋅ + ⋅ ⋅ + + ⋅ + − = ⇒

( ) ( ) ( )3 2 2

2 15 75 125 7 10 25 7 5 2 0x x x x x x⇒ ⋅ + ⋅ + ⋅ + − ⋅ + ⋅ + + ⋅ + − = ⇒

3 2 22 30 150 250 7 70 175 7 35 2 0x x x x x x⇒ ⋅ + ⋅ + ⋅ + − ⋅ − ⋅ − + ⋅ + − = ⇒

3 23 2 2 23 87 108 0

2 23 87 108 0 .2 , 23 , 87 , 108

x x xx x x

a b c d

⋅ + ⋅ + ⋅ + =⇒ ⋅ + ⋅ + ⋅ + = ⇒

= = = =

Sada je:

23.

1 2 3 1 2 3 2

bx x x x x x

a+ + = − ⇒ + + = −

Odgovor je pod D.

3.inačica

( ) ( ) ( ) ( ) ( ) ( )3 2 3 2

2 5 7 5 7 5 2 0 2 5 2 7 5 7 5 0x x x x x x⋅ + − ⋅ + + ⋅ + − = ⇒ ⋅ + − − ⋅ + + ⋅ + = ⇒

( )( ) ( ) ( )( )3

2 5 1 7 5 5 1 0x x x⇒ ⋅ + − − ⋅ + ⋅ + − = ⇒

3

( )( ) ( ) ( )( ) ( ) ( )2 2

2 5 1 5 5 1 1 7 5 5 1 0x x x x x⇒ ⋅ + − ⋅ + + + ⋅ + − ⋅ + ⋅ + − = ⇒

( ) ( )( ) ( ) ( )2

2 5 1 5 5 1 7 5 4 0x x x x x⇒ ⋅ + − ⋅ + + + + − ⋅ + ⋅ + = ⇒

( ) ( ) ( ) ( )2

2 4 10 25 5 1 7 5 4 0x x x x x x⇒ ⋅ + ⋅ + ⋅ + + + + − ⋅ + ⋅ + = ⇒

( ) ( ) ( ) ( )2

2 4 11 31 7 5 4 0x x x x x⇒ ⋅ + ⋅ + ⋅ + − ⋅ + ⋅ + = ⇒

( ) ( ) ( )( )24 2 11 31 7 5 0x x x x⇒ + ⋅ ⋅ + ⋅ + − ⋅ + = ⇒

( ) ( ) ( ) ( )2 24 2 22 62 7 35 0 4 2 15 27 0x x x x x x x⇒ + ⋅ ⋅ + ⋅ + − ⋅ − = ⇒ + ⋅ ⋅ + ⋅ + = ⇒

4 0.

22 15 27

x

x x

+ =⇒

⋅ + ⋅ +

• Riješimo prvu linearnu jednadžbu.

4 0 4.1

x x+ = ⇒ = −

• Riješimo drugu kvadratnu jednadžbu.

2 , 15 , 272

2 2 15 27 022 15 27 0

42 , 15 , 27

2,3 2

a b c

x xx x

b b a ca b c x

a

= = =

⋅ + ⋅ + =⋅ + ⋅ + = ⇒ ⇒ ⇒

− ± − ⋅ ⋅= = = =

⋅

215 15 4 2 27 15 225 216 15 9

2,3 2,3 2,32 2 4 4x x x

− ± − ⋅ ⋅ − ± − − ±⇒ = ⇒ = ⇒ = ⇒

⋅

15 3 123

22 2 215 3 4 4.92,3 15 3 184

12

4

18

43 23 3 34 4

xx x x

xx

x x x

− += −= = − = −

− ±⇒ = ⇒ ⇒ ⇒ ⇒

− − = −= = − = −

Zbroj rješenja jednadžbe iznosi:

( )9 9 4 3 9

4 3 4 31 2 3 1 2 3 1 2 32 2 1 1 2

x x x x x x x x x+ + = − + − + − ⇒ + + = − − − ⇒ + + = − − − ⇒

8 6 9 23.

1 2 3 1 2 32 2x x x x x x

− − −⇒ + + = ⇒ + + = −

Odgovor je pod D.

Vježba 181

Koliko iznosi umnožak rješenja jednadžbe ( ) ( ) ( )3 2

2 5 7 5 7 5 2 0?x x x⋅ + − ⋅ + + ⋅ + − =

. 54 . 54 . 108 . 108A B C D− −

Rezultat: B.

Zadatak 182 (Marljivi dečki, Magnusgymnasium)

Riješi jednadžbu 2

2 3 2 0.x x⋅ − ⋅ − =

Rješenje 182

4

Ponovimo!

Za realni broj x njegova je apsolutna vrijednost (modul) broj │x│ koji određujemo na ovaj način:

, .

, 0

0

x xx

x x

≥=

− <

Ako je broj x pozitivan ili nula, tada je on jednak svojoj apsolutnoj vrijednosti. Za svaki x, x ≥ 0,

vrijedi │x│= x.

Ako je x negativan broj, njegova apsolutna vrijednost je suprotan broj – x koji je pozitivan. Za svaki x,

x < 0, je │x│= – x.


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Zadanu kvadratnu jednadžbu preoblikujemo u dvije: jednu u slučaju kada je x ≥ 0, a drugu za x < 0.

• slučaj x ≥ 0

20 ,2 2 3 2 0

2 3 2 02

2 , 3 , 22 3 2 0

x x xx x

x xa b cx x

≥ =⋅ − ⋅ − =

⇒ ⋅ − ⋅ − = ⇒ ⇒= = − = −⋅ − ⋅ − =

2 , 3 , 2

24

1,2 2

a b c

b b a cx

a

= = − = −

⇒ ⇒− ± − ⋅ ⋅

=⋅

( ) ( ) ( )2

3 3 4 2 2 3 9 16 3 25

1,2 1,2 1,22 2 4 4x x x

− − ± − − ⋅ ⋅ − ± + ±⇒ = ⇒ = ⇒ = ⇒

⋅

3 5 82

11 1 13 5 4 411,2 3 5 24

2 22 2 24 4

8

4

2

4

xx x x

xx

x x x

+== = =

±⇒ = ⇒ ⇒ ⇒ ⇒ ⇒

− = −= = − = −

21uvjet

0 nema sm2.1 1

2isla

2

x

xxx

=

⇒ ⇒ ⇒ == −≥

• slučaj x < 0

( )0 ,

2 22 3 2 0 2 3 2 0

22 3 2 0

x x xx x x x

x x

< = −

⇒ ⋅ − ⋅ − − = ⇒ ⋅ + ⋅ − = ⇒⋅ − ⋅ − =

2 , 3 , 22

2 3 2 02

42 , 3 , 2

1,2 2

a b c

x xb b a c

a b c xa

= = = −

⋅ + ⋅ − =⇒ ⇒ ⇒

− ± − ⋅ ⋅= = = − =

⋅

( )2

3 3 4 2 2 3 9 16 3 25

1,2 1,2 1,22 2 4 4x x x

− ± − ⋅ ⋅ − − ± + − ±⇒ = ⇒ = ⇒ = ⇒

⋅

5

3 5 21

1 1 13 5 4 4 1 21,2 3 5 84

222 2 24 4

2

4

8

4

x x xx

x

xx x x

− += = =

=− ±⇒ = ⇒ ⇒ ⇒ ⇒ ⇒

− −= −= = − = −

nema smislauvjet

0

1

1 2 2.2

22

x

xx

x

=⇒ ⇒ ⇒ =

= −<

−

Vježba 182

Riješi jednadžbu 2

2 3 2.x x⋅ − ⋅ =

Rezultat: x1 = 2, x2 = – 2. Zadatak 183 (4A, 4B, TUPŠ)

Riješite nejednadžbu ( ) ( )2

2 1 3 2 1 2 0x x⋅ − + ⋅ ⋅ − + > i napišite rješenja uz pomoć intervala.

Rješenje 183

Ponovimo!

Metoda testiranja točaka

Kako odrediti je li kvadratna funkcija 2

y a x b x c= ⋅ + ⋅ + pozitivna, negativna ili nula?

• Najprije odredimo njezine realne nultočke tako da riješimo kvadratnu jednadžbu

20.a x b x c⋅ + ⋅ + =

• Nultočke dijele brojevni pravac (skup realnih brojeva, R) na intervale.

• Iz svakog intervala izaberemo po jedan x i uvrstimo ga u 2

0.a x b x c⋅ + ⋅ + = Predznak

dobivene vrijednosti određuje predznak za cijeli interval.

• Intervali koji imaju traženi predznak rješenje su problema.

( ) ( )2 2 2

2 , ., 0,a bn n n

a b a a b b a b a b a b cc c

− = − ⋅ ⋅ + ⋅ = ⋅ > > ⇒ >

Za broj x0 kažemo da je nultočka (korijen) funkcije f ako vrijedi

( ) .00

f x =


0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Neka je U univerzalni skup, te A i B proizvoljni skupovi koji su podskupovi skupa U. Tada je:

A B∪ unija skupova A i B,

A∪ B = { }:x U x A ili x B∈ ∈ ∈ unija skupova A i B

Unija dva ili više skupa je skup koji se sastoji od svih elemenata zadanih skupova.


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

.a c a d b c

b d b d

⋅ − ⋅− =

⋅

6

Preoblikujemo zadanu nejednadžbu.

( ) ( )2 2

2 1 3 2 1 2 0 4 4 1 6 3 2 0x x x x x⋅ − + ⋅ ⋅ − + > ⇒ ⋅ − ⋅ + + ⋅ − + > ⇒

12 2 2

4 4 6 0 4 4 6 0 2 02 43x x x x x x x x⇒ ⋅ − ⋅ + ⋅ > ⇒ ⋅ − ⋅ + ⋅ > ⇒ ⋅ + ⋅+ − >+ ⇒

/ : 22 2

4 2 0 2 0.x x x x⇒ ⋅ + ⋅ > ⇒ ⋅ + >

Zatim nađemo nultočke kvadratne jednadžbe.

( )002 12 0 2 1 0

2 1 0 2 1

xxx x x x

x x

==⋅ + = ⇒ ⋅ ⋅ + = ⇒ ⇒ ⇒

⋅ + = ⋅ = −

/ : 2

010

1 .12 1

2 2

xx

x x

==

⇒ ⇒⋅ = − = −

Nultočke ucrtamo na os x.

0-

1

2

+ ∞∞∞∞- ∞∞∞∞

EF

U točkama 1

i 02

− vrijedi jednakost =, pa one nisu rješenja naše stroge nejednakosti >. Zato ih nismo

popunili.

• Odaberemo x manji od 1

,2

− na primjer x = – 1 i uvrstimo ga u 2

2 .y x x= ⋅ +

( ) ( )1 2

2 1 1 2 1 1 2 1 1 .2

20

xy y y y

y x x

= −⇒ = ⋅ − + − ⇒ = ⋅ − ⇒ = − ⇒ =

= ⋅>

+

Rezultat je pozitivan što znači da je 2

2y x x= ⋅ + pozitivno na cijelom intervalu 1

, .2

− ∞ −

Upišimo + iznad tog intervala 1

, .2

− ∞ −

+

0-

1

2

+ ∞∞∞∞- ∞∞∞∞

EF

• Odaberemo x između 1

i 0,2

− na primjer 1

4x = − i uvrstimo ga u

22 .y x x= ⋅ +

1 21 1 1 1 1 1

4 2 24 4 16 4

216 42

2

xy y y

y x x

= −⇒ = ⋅ − + − ⇒ = ⋅ − ⇒ = ⋅ − ⇒

= ⋅ +

1 1 1 2 1.

8 4 80

8y y y

−⇒ = − ⇒ = ⇒ <= −

Rezultat je negativan što znači da je 2

2y x x= ⋅ + negativan na cijelom intervalu 1

, 0 .2

−

7

Upišimo - iznad tog intervala1

, 0 .2

−

-+

0-

1

2

+ ∞∞∞∞- ∞∞∞∞

EF

• Odaberemo x veći od 0, na primjer x = 1 i uvrstimo ga u 2

2 .y x x= ⋅ +

1 22 1 1 2 1 1 2 3 01 .

22

xy y y y

y x x

=⇒ = ⋅ + ⇒ = ⋅ + ⇒ = + =

= ⋅>⇒

+

Rezultat je pozitivan što znači da je 2

2y x x= ⋅ + pozitivno na cijelom intervalu 0, .+ ∞

Upišimo + iznad tog intervala 0, .+ ∞

+-+

0-

1

2

+ ∞∞∞∞- ∞∞∞∞

EF

Nejednadžba

22 0,x x⋅ + >

odnosno polazna nejednadžbe

( ) ( )2

2 1 3 2 1 2 0x x⋅ − + ⋅ ⋅ − + >

vrijedi za

1i 0.

2x x< − >

Zapis rješenje uz pomoć intervala glasi:

1, 0, .

2x ∈ − ∞ − + ∞∪

+-

+

0-

1

2

+ ∞∞∞∞- ∞∞∞∞

EC

Vježba 183

Riješite nejednadžbu ( ) ( )2

2 1 3 2 1 2 0x x⋅ − + ⋅ ⋅ − + < i napišite rješenja uz pomoć intervala.

Rezultat: 1

, 0 .2

−

Zadatak 184 (Zvone, tehnička škola)

Odredite sva realna rješenja jednadžbe 4

5 135 0.y y⋅ − ⋅ =

Rješenje 184

Ponovimo!

8

( )1

:, , .n n n n m n m

a b a b a a a a a−

⋅ = ⋅ = =

( ) ( )3 3.

2 2a b a b a a b b− = − ⋅ + ⋅ +


0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Diskriminanta kvadratne jednadžbe

20a x b x c⋅ + ⋅ + =

je broj

24 .D b a c= − ⋅ ⋅

• Ako je D > 0, jednadžba ima dva realna rješenja.

• Ako je D = 0, jednadžba ima jedno dvostruko realno rješenje.

• Ako je D < 0, jednadžba ima kompleksno – konjugirana rješenja.

( )/ :4 4 4 3

5 135 0 5 135 0 27 0 1 275 0y y y y y y y y⋅ − ⋅ = ⇒ ⋅ − ⋅ = ⇒ − ⋅ = ⇒ ⋅ − ⋅ = ⇒

( )( ) ( ) ( )( )3 21 3 0 1 3 1 3 3 0y y y y y y⇒ ⋅ − ⋅ = ⇒ ⋅ − ⋅ ⋅ + ⋅ + ⋅ = ⇒

( ) ( ) ( ) ( )2 21 3 1 3 9 0 1 3 9 3 1 0y y y y y y y y⇒ ⋅ − ⋅ ⋅ + ⋅ + ⋅ = ⇒ ⋅ − ⋅ ⋅ ⋅ + ⋅ + = ⇒

0

1 3 0 .

29 3 1 0

y

y

y y

=

⇒ − ⋅ =

⋅ + ⋅ + =

Prva jednadžba

rješenje je0 0 .1

realnoy y= ⇒ =

Druga jednadžba

( )/ :1

1 3 0 3 1 3 3 rješenje je realno1 .3

y y y y−− ⋅ = ⇒ − ⋅ = − ⇒ − ⋅ = − ⇒ =

Treća jednadžba

2 9 , 3 , 12 29 3 1 09 3 1 0 3 4 9 1

29 , 3 , 1 4

a b cy yy y D

a b c D b a c

= = =⋅ + ⋅ + =

⋅ + ⋅ + = ⇒ ⇒ ⇒ = − ⋅ ⋅ ⇒= = = = − ⋅ ⋅

0 rješenja nisu realna već konjugirano kompl9 36 27 eksni bro vi.jeD D⇒ = − ⇒ = <−

Vježba 184

Odredite sva realna rješenja jednadžbe 4

2 16 0.y y⋅ − ⋅ =

Rezultat: 1

0, .2

Zadatak 185 (Ante, tehnička škola)

Kvadratna jednadžba x2 + b · x + c = 0 ima dvostruko rješenje x1 = x2 = – 5. Koliki je

koeficijent b te kvadratne jednadžbe?

Rješenje 185

Ponovimo!

9

( )21 2 2

, .2,n m n m

a a a a a a b a a b b+

= ⋅ = + = + ⋅ ⋅ +

( )22 2

2 , 0, .0n

a a b b a b a n a− ⋅ ⋅ + = − ≠ ⇒ =

Polinom drugog stupnja (kvadratna funkcija) ima oblik

( )2

,f x a x b x c= ⋅ + ⋅ +

gdje su a ≠ 0, b i c realni brojevi. Broj a naziva se vodeći koeficijent polinoma drugog stupnja, b

linearni koeficijent, a c slobodni koeficijent polinoma.

Poučak o jednakosti polinoma:

Dva polinoma jednaka su ako i samo ako su istog stupnja i ako su im koeficijenti uz iste potencije

jednaki.

Faktorizacija kvadratnog trinoma

Svaki se kvadratni trinom može napisati u obliku

( ) ( )21 2

,a x b x c a x x x x⋅ + ⋅ + = ⋅ − ⋅ −

gdje su x1 i x2 rješenja pripadne kvadratne jednadžbe a · x2 + b · x + c = 0.

Graf kvadratne funkcije

( )2

, 0f x a x b x c a= ⋅ + ⋅ + ≠

je parabola

2.y a x b x c= ⋅ + ⋅ +

Tjeme T najniža je točka parabole i parabola je otvorena prema gore ako je a > 0.

Tjeme T najviša je točka parabole i parabola je otvorena prema dolje ako je a < 0.

Za broj x0 kažemo da je nultočka (korijen) funkcije f ako vrijedi

( ) .00

f x =

Kvadratna funkcija (polinom drugog stupnja)

( )2

,f x a x b x c= ⋅ + ⋅ +

ima ekstrem u točki s apscisom

0 2.

bx

a= −

⋅

Ekstrem je minimum ako je a > 0, maksimum ako je a < 0.


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


20a x b x c⋅ + ⋅ + =

je broj

24 .D b a c= − ⋅ ⋅

• Ako je D > 0, jednadžba ima dva realna rješenja.



1.inačica

Kvadratna jednadžba x2 + b · x + c = 0 ima dvostruko rješenje x1 = x2 = – 5 pa vrijedi rastav.

( ) ( ) 51

21 22

x b x c x x x x x x+ ⋅ = = −+ = − ⋅ − ⇒ ⇒

( )( ) ( )( ) ( ) ( )2 2

5 5 5 5x b x c x x x b x c x x+ ⋅ + = − − ⋅ − − ⇒ + ⋅ + = + ⋅ + ⇒

10

( )jednakost

polino

22 2 25 10 2

ma5x b x c x x b x c x x⇒ + ⋅ + = + ⇒ + ⋅ + = + ⋅ + ⇒ ⇒

1010 .

25rješenje

bb

c

=⇒ ⇒ =

=

2.inačica

Pripadna kvadratna funkcija dana je pravilom

( )( )

22

.1 , ,

f x x b x cf x x b x c

a b b c c

= + ⋅ += + ⋅ + ⇒

= = =

Njezin graf je parabola okrenuta otvorom uvis jer je a = 1 > 0.

6

5,5

5

4,5

4

3,5

3

2,5

2

1,5

1

0,5

-9 -8 -7 -6 -5 -4 -3 -2 -1

- 5

y

x

A

Budući da kvadratna jednadžba ima dvostruko rješenje x1 = x2 = – 5, kvadratna funkcija ima dvostruku

nultočku x0 = – 5 pa apscisa tjemena glasi x0 = – 5.

Dalje imamo:

15 5 5 5 10.

5 2 1 2 2 2/ 2

0 20

a b b b b

ab

x

bx

=⇒ ⇒ − == − ⋅

⋅− ⇒ − = − ⇒ = ⇒ = ⇒ =

= − ⋅

3.inačica

Odredimo koeficijente kvadratne jednadžbe.

22 0

0 .1 , ,

x b x cx b x c

a b b c c

+ ⋅ + =+ ⋅ + = ⇒

= = =

Budući da je x = – 5 rješenje jednadžbe, uvrstit ćemo u nju tu vrijednost.

( ) ( )2

205 5 0 25 5 0 5 25.

5

x b x cb c b c c b

x

+ ⋅ + =⇒ − + ⋅ − + = ⇒ − ⋅ + = ⇒ = ⋅ −

= −

Kvadratna jednadžba ima dvostruko realno rješenje pa njezina diskriminanta mora biti jednaka nuli.

22 24 0

4 1 0 4 0.1

b a cb c b c

a

− ⋅ ⋅ =⇒ − ⋅ ⋅ = ⇒ − ⋅ =

=

Iz sustava jednadžbi dobijemo vrijednost koeficijenta b.

11

( )metoda

zamjene

5 25 2 24 5 25 0 20 100 0

24 0

c bb b b b

b c

= ⋅ −⇒ ⇒ − ⋅ ⋅ − = ⇒ − ⋅ + = ⇒

− ⋅ =

( )2

10 0 10 0 10.b b b⇒ − = ⇒ − = ⇒ =

Vježba 185

Kvadratna jednadžba x2 + b · x + c = 0 ima dvostruko rješenje x1 = x2 = – 4. Koliki je

koeficijent b te kvadratne jednadžbe?

Rezultat: b = 8.

Zadatak 186 (Ispravak zadatka)

Riješite jednadžbu: ( ) ( ) ( ) ( ) ( )2 2 2 2 2

1 2 3 4 ... 11 385.x x x x x+ + + + + + + + + + =

U zbirci je zadatak napisan kao ( ) ( ) ( ) ( ) ( )2 2 2 2 2

1 2 3 4 .. 3. 1 51 .x x x x x+ + + + + + + + + + =

Zahvaljujem kolegi Tiboru na uočenoj pogrešci.

Rješenje 186

Ponovimo!

( )( )12 2 2

2 1 2, 3 4 ...2

.n n

a b a a b b n⋅ +

+ = + ⋅ ⋅ + + + + + + =

( ) ( )1 2 12 2 2 2 21 2 3 4 ...

6.

n n nn

⋅ + ⋅ ⋅ ++ + + + + =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

( ) ( ) ( ) ( ) ( )2 2 2 2 2

1 2 3 4 ... 11 385x x x x x+ + + + + + + + + + = ⇒

2 2 2 2 22 1 4 4 6 9 8 16 ... 22 121 385x x x x x x x x x x⇒ + ⋅ + + + ⋅ + + + ⋅ + + + ⋅ + + + + ⋅ + = ⇒

( ) ( )2

11 2 4 6 8 ... 22 1 4 9 16 ... 121 385x x x x x x⇒ ⋅ + ⋅ + ⋅ + ⋅ + ⋅ + + ⋅ + + + + + + = ⇒

( ) ( )2 2 2 2 2 211 2 1 2 3 4 ... 11 1 2 3 4 ... 11 385x x⇒ ⋅ + ⋅ ⋅ + + + + + + + + + + + = ⇒

( ) ( ) ( )11 11 1 11 11 1 2 11 1211 2 385

2 6x x

⋅ + ⋅ + ⋅ ⋅ +⇒ ⋅ + ⋅ ⋅ + = ⇒

11 12 11 12 23 11 12 11 232 211 2 385 11 385

122

2 62 6x x x x

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⇒ ⋅ + ⋅ ⋅ + = ⇒ ⋅ + ⋅ ⋅ + = ⇒

2 211 11 12 11 2 23 385 11 132 506 385x x x x⇒ ⋅ + ⋅ ⋅ + ⋅ ⋅ = ⇒ ⋅ + ⋅ + = ⇒

2 211 132 506 385 0 11 132 121 0x x x x⇒ ⋅ + ⋅ + − = ⇒ ⋅ + ⋅ + = ⇒

22 2 12 11 0

11 132 121 0 12/ : 11 01 , 1

12 , 11

1x x

x x x xa b c

+ ⋅ + =⇒ ⋅ + ⋅ + = ⇒ + ⋅ + = ⇒ ⇒

= = =

12

1 , 12 , 112

12 12 4 1 11 12 144 442

4 1,2 1,22 1 21,2 2

a b c

x xb b a c

xa

= = =− ± − ⋅ ⋅ − ± −

⇒ ⇒ = ⇒ = ⇒− ± − ⋅ ⋅ ⋅=

⋅

12 10 2

1 112 100 12 10 2 21,2 1,2 12 10 222 2

2 22 2

x x

x x

x x

− += = −

− ± − ±⇒ = ⇒ = ⇒ ⇒ ⇒

− −= = −

2

2

22

2

11 1.

112

2

x x

xx

= − = −⇒ ⇒

= −= −

Vježba 186

Riješite jednadžbu: ( ) ( ) ( )2 2 2

1 2 3 5.x x x+ + + + + =

Rezultat: 1, 3.1 2

x x= − = −

Zadatak 187 (Lara, gimnazija)

Automobil i autobus krenuli su istodobno na put dug 360 km. Automobil je u prosjeku vozio

15 km / h brže od autobusa pa je stigao 48 minuta ranije. Kolika je bila prosječna brzina autobusa?

Rješenje 187

Ponovimo!

11 60 min 1, .min

60h h= =

Jednoliko pravocrtno gibanje duž puta s jest gibanje pri kojem vrijedi izraz

,s

s v t tv

= ⋅ ⇒ =

gdje je v stalna, konstantna brzina kojom se tijelo giba.


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

48 4360 , , 15 / , 48 min

60

48

60 5s km v v v km h t h h hab b

= = + ∆ = = = =

Vremena za koja automobil i autobus prijeđu put s iznose:

• automobil

15

s st ta a

v va b

= ⇒ =+

• autobus

.s

tb v

b

=

Budući da je automobil stigao ∆t vremena ranije od autobusa, vrijedi:

13

4 360 360 4

15 5 15 5

s st t tab v v v v

b b b b

− = ∆ ⇒ − = ⇒ − = ⇒+ +

( )( ) ( )

5 15/

4

360 360 4450 15 450 15

15 5v v v vb b b bv v

b b

v vb b⇒ − = ⇒ ⋅ + − ⋅ = ⋅

⋅ ⋅ ++⋅ ⇒

+

2 2450 6 750 450 15 450 6 750 1450 5v v v v v v

b b b b b bv vb b

⇒ ⋅ + − ⋅ = + ⋅ ⇒ = +− ⋅+ ⋅⋅ ⇒

2 2 26 750 15 15 6 750 15 6750 0v v v v v v

b b b b b b⇒ = + ⋅ ⇒ + ⋅ = ⇒ + ⋅ − = ⇒

( )

1 , 15 , 67502

15 6 750 02

41 , 15 , 6 750

1,2 2

a b cv vb b

b b a cva b cb a

= = = −+ ⋅ − =

⇒ ⇒ ⇒− ± − ⋅ ⋅== = = −

⋅

( )( )

( )2

15 15 4 1 6 750 15 225 27 000

1,2 1,22 1 2v vb b

− ± − ⋅ ⋅ − − ± +⇒ = ⇒ = ⇒

⋅

( ) ( )( )

( )

15 165

115 27 225 15 165 2

1,2 1,2 15 1652 2

2 2

vb

v vb b

vb

− +=

− ± − ±⇒ = ⇒ = ⇒ ⇒

− −=

( )

( )

( )

( )

( )

( )

150

2

180 nema smisla

2

150751 12 1

75 .180 90

22 22

v v vb b b kmvb hvv v bb b

= = =

⇒ ⇒ ⇒ ⇒ =

= −= − = −

Prosječna brzina autobusa bila je 75 km / h.

v + 15v

.

Vježba 187

Automobil i autobus krenuli su istodobno na put dug 360 km. Autobus je u prosjeku vozio

15 km / h sporije pa je automobil stigao 48 minuta ranije. Kolika je bila prosječna brzina autobusa?

Rezultat: 75 km / h.

Zadatak 188 (Lara, gimnazija)

Lara ☺ do posla vozi 45 km. Ako poveća prosječnu brzinu za 5 km / h, stići će 6 minuta ranije

nego obično. Odredite kojom prosječnom brzinom obično vozi na posao.

Rješenje 188

Ponovimo!

11 60 min 1, .min

60h h= =

Jednoliko pravocrtno gibanje duž puta s jest gibanje pri kojem vrijedi izraz

,s

s v t tv

= ⋅ ⇒ =

gdje je v stalna, konstantna brzina kojom se tijelo giba.

14


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

6 145 , , 5 / , 6 min

1 2 1 60 10

6

60s km v v v km h t h h h= = + ∆ = = = =

Vremena za koja Lara vozi sporije ili brže na putu s iznose:

• sporija vožnja brzinom v1

11

st

v=

• brža vožnja brzinom v2

.2 2 5

2 1

s st t

v v= ⇒ =

+

Budući da je bržom vožnjom stigla ∆t vremena ranije nego obično, vrijedi:

45 45 1

1 2 5 5 101 1 1 1

s st t t t

v v v v− = ∆ ⇒ − = ∆ ⇒ − = ⇒

+ +

( ) ( ) ( )/ 1045 45 1

450 5 450 51 1 1 15 10 1

11

1

5v vv v v vv v

⋅ ⋅ ⋅ +⇒ − = ⇒ ⋅ + − ⋅ = ⋅ + ⇒+

2 2450 2 250 450 5 450 2 250 5

1 1 1 1 1 1450

1 1v v v v v vv v⇒ ⋅ + − ⋅ = + ⋅ ⇒ =⋅ +−+ ⋅⋅ ⇒

2 2 22 250 5 5 2 250 5 2 250 0

1 1 1 1 1 1v v v v v v⇒ = + ⋅ ⇒ + ⋅ = ⇒ + ⋅ − = ⇒

( )

1 , 5 , 2 2502

5 2 250 01 1 2

41 , 5 , 2 250

1 1,2 2

a b cv v

b b a ca b c v

a

= = = −+ ⋅ − =

⇒ ⇒ ⇒− ± − ⋅ ⋅= = = − =

⋅

( )( )

( )2

5 5 4 1 2 250 5 25 9 000

1 11,2 1,22 1 2v v

− ± − ⋅ ⋅ − − ± +⇒ = ⇒ = ⇒

⋅

( ) ( )( )

( )

5 95

1 15 9 025 5 95 21 11,2 1,2 5 952 2

1 2 2

v

v v

v

− +=

− ± − ±⇒ = ⇒ = ⇒ ⇒

− −=

( )

( )

( )

( )

( )

( )

90451 11 1 12 1

45 .1100 50

1 21 1

90

2

100 nema smisl2 222

a

v v vkm

vhvv v

= = =

⇒ ⇒ ⇒ ⇒ =

= −= − = −

Prosječna brzina je obično 45 km / h.

15

Vježba 188

Ana do posla vozi 45 km. Ako poveća prosječnu brzinu za 5 km / h, stići će 6 minuta ranije

nego obično. Odredite iznos veće brzine kojom vozi na posao.

Rezultat: 50 km / h.

Zadatak 189 (Vibrato et pizzicato ☺☺☺☺, gimnazija)

Riješi jednadžbu: ( ) ( )2 2 2

.x a x a b b x a b− ⋅ − ⋅ = ⋅ + ⋅

Rješenje 189

Ponovimo!

( ),1

.0, ,,2mn m n m n n m

a a a a a a a a a a+ ⋅

= ⋅ = = = ≥

( ) ( ),2 22 2 2

2 .a a a b a a b b− = + = + ⋅ ⋅ +

( ) .2 2 2 2

2 2 2a b c a b c a b a c b c+ + = + + + ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) ( )2 2 2 2 2 3 2 3

x a x a b b x a b x a x a b b x a b− ⋅ − ⋅ = ⋅ + ⋅ ⇒ − ⋅ + ⋅ = ⋅ + ⋅ ⇒

( )2 2 3 2 3 2 2 2 3 30 0x a x a b b x a b x a b x a b a b⇒ − ⋅ + ⋅ − ⋅ − ⋅ = ⇒ − + ⋅ + ⋅ − ⋅ = ⇒

( )

( )

2 2 2 3 30

2 2 3 31 , ,

x a b x a b a b

a b a b c a b a b

− + ⋅ + ⋅ − ⋅ =

⇒ ⇒

= = − + = ⋅ − ⋅

( )2 2 3 31 , ,

24

1,2 2

a b a b c a b a b

b b a cx

a

= = − + = ⋅ − ⋅

⇒ ⇒− ± − ⋅ ⋅

=⋅

( )( ) ( )( ) ( )2

2 2 2 2 3 34 1

1,2 2 1

a b a b a b a b

x

− − + ± − + − ⋅ ⋅ ⋅ − ⋅

⇒ = ⇒⋅

( ) ( )2

2 2 2 2 3 34

1,2 2

a b a b a b a b

x

+ ± + − ⋅ ⋅ − ⋅

⇒ = ⇒

2 2 4 2 2 4 3 32 4 4

1,2 2

a b a a b b a b a bx

+ ± + ⋅ ⋅ + − ⋅ ⋅ + ⋅ ⋅⇒ = ⇒

2 2 4 2 2 4 3 2 2 34 4 2 4

1,2 2

a b a a b b a b a b a bx

+ ± + ⋅ ⋅ + − ⋅ ⋅ − ⋅ ⋅ + ⋅ ⋅⇒ = ⇒

16

( ) ( )2

2 2 2 2 2 2 2 22 2

1,2 1,22 2

a b a a b b a b a a b b

x x

+ ± − ⋅ ⋅ − + ± − ⋅ ⋅ −

⇒ = ⇒ = ⇒

( )

( )

2 2 2 22 2 2 22

2

1 12 2

2 2 2 22 2 2 222

2 22 2

a b a a b ba b a a b b

x x

a b a a b ba b a a b bx

x

+ + − ⋅ ⋅ −+ + − ⋅ ⋅ −

= =

⇒ ⇒ ⇒

+ − + ⋅ ⋅ ++ − − ⋅ ⋅ −=

=

2 2 2 22 2

1 12 2

2 2 2 22 2

2 22 2

2 2

2 2

a a a b a a a bx x

b a b b

b b

b a b bx

ax

a

+ − ⋅ ⋅ + − ⋅ ⋅= =

⇒ ⇒ ⇒

+ −

−+ + ⋅ ⋅ + + ⋅ ⋅ += =

( )

( )

( )

( )

222 2

1 1 12 2

2 22 22 22 2

2

2

22

2

a a b a a ba a bx x x

b a b b a ba b b x xx

⋅ ⋅ − ⋅ ⋅ −⋅ − ⋅ ⋅= = =

⇒ ⇒ ⇒ ⇒⋅ ⋅ + ⋅ ⋅ +⋅ ⋅ + ⋅ = ==

( )

( )

1.

2

x a a b

x b a b

= ⋅ −⇒

= ⋅ +

Vježba 189

Riješi jednadžbu: ( ) ( )2 2 2

.x a a b x b x a b+ ⋅ ⋅ − = ⋅ + ⋅

Rezultat: ( ) ( ), .1 2

x a a b x b a b= ⋅ − = ⋅ +

Zadatak 190 (Vibrato et pizzicato ☺☺☺☺, gimnazija)

Riješi jednadžbu:

2 2 2

.2

x b x a b a x x a

x a b xx a x b x a b

+ + − − ⋅ +− =

− −− ⋅ − ⋅ + ⋅

Rješenje 190

Ponovimo!

( ) ( )2 2 2

0, ., ,a b a b a b a a a a b a b+ ⋅ − = − = ≥ ⋅ = ⋅


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

metoda

grupir

2 2

2 a ja

2

n

x b x a b a x x a


+ + − − ⋅ +− = ⇒ ⇒

− −− ⋅ − ⋅ + ⋅

17

( ) ( )

2 2 2

2

x b x a b a x x a


+ + − − ⋅ +⇒ − = ⇒

− −− ⋅ + − ⋅ + ⋅

( ) ( ) ( ) ( )

2 2 2 2 2 2

x a x a

x b x a b a x x a x b x a b a x x a

x a x x a b x a b x x a x b b x

+ + − − ⋅ + + + − − ⋅ +⇒ − = ⇒ − = ⇒

− ⋅ − − ⋅ − ⋅ −− − − ⋅ −−

( ) ( ) ( ) ( )

2 2 2 2 2 2

0x b x a b a x x a x b x a b a x x a

x a x a x b b x x a x a x b b x

+ + − − ⋅ + + + − − ⋅ +⇒ − = ⇒ − − = ⇒

− − ⋅ − − − − ⋅ − −

( ) ( )

2 2 2

0.x b x a b a x x a

x a x a x b x b

+ + − − ⋅ +⇒ − + =

− − ⋅ − −

Rasprava! Budući da nazivnik ne smije biti jednak nuli, slijedi:

0.

0

x a x a

x b x b

− ≠ ≠⇒

− ≠ ≠

Dalje pišemo:

( ) ( )

2 2 2

0x b x a b a x x a

x a x a x b x b

+ + − − ⋅ +− + = ⇒

− − ⋅ − −

( ) ( )( ) ( )

2 2

/

2

0x b x a b a x x a

x a x a x bx a

x bx b

+ + − − ⋅ +⇒ − + = ⇒

− − ⋅⋅ − ⋅ −

− −

( ) ( ) ( ) ( ) ( )2 2 2

0x b x b x a b a x x a x a⇒ + ⋅ − − + − − ⋅ + + ⋅ − = ⇒

2 2 2 2 2 2 20x b x a b a x x a⇒ − − − + + ⋅ + − = ⇒

2 2 2 22 2 2 2 2 20 0a a x x a a a x x ax b x b⇒ − + ⋅ + − = ⇒ − + ⋅ + −− + =− ⇒

2 22 02 2

2 02

1 , , 2

x a x ax a x a

a b a c a

+ ⋅ − ⋅ =⇒ + ⋅ − ⋅ = ⇒ ⇒

= = = − ⋅

( )2 2 21 , , 2 4 1 2

2 1,24 2 11,2 2

a b a c a a a a

xb b a c

xa

= = = − ⋅ − ± − ⋅ ⋅ − ⋅

⇒ ⇒ = ⇒− ± − ⋅ ⋅ ⋅

=⋅

2 2 28 9 3

1,2 1,2 1,22 2 2

a a a a a a ax x x

− ± + ⋅ − ± ⋅ − ± ⋅⇒ = ⇒ = ⇒ = ⇒

3 2

1 1 12 2

3 4

2 2 22 2

2

2

4

2

a a a ax x x

a a a ax x x

− + ⋅ ⋅ ⋅= = =

⇒ ⇒ ⇒ ⇒− − ⋅ ⋅ ⋅

= = − = −

18

nije rješenje zbog rasp1

2 .2

2

ravex ax a

x a

=⇒ ⇒ = − ⋅

= − ⋅

Vježba 190

Riješi jednadžbu:

2 2 2

.2

x b x a x a b a x

x a b x x a x b x a b

+ + + − − ⋅= +

− − − ⋅ − ⋅ + ⋅

Rezultat: 2 .x a= − ⋅

Zadatak 191 (2B, TUPŠ)

Riješi jednadžbu pri čemu su a i b realni brojevi: ( )2 2 2 20.a b x a b x a⋅ ⋅ − + ⋅ + =

Rješenje 191

Ponovimo!

( ) ( ), , ,2 2 1

.mn m n m n n m

a a a a a a a a a+ ⋅

− = = ⋅ = =

( ) ( )2 22 2 2 2 2

2 2, , ., 0a b a a b b a a b b a b a a a+ = + ⋅ ⋅ + − ⋅ ⋅ + = − = ≥

.

na n m

ama

−=


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( )( )

( )

2 2 2 20

2 2 2 20

2 2 2, ,

a b x a b x a

a b x a b x a

a a b b a b c a

⋅ ⋅ − + ⋅ + =

⋅ ⋅ − + ⋅ + = ⇒ ⇒

= ⋅ = − + =

( )2 2 2, ,

24

1,2 2

a a b b a b c a

b b a cx

a

= ⋅ = − + =

⇒ ⇒− ± − ⋅ ⋅

=⋅

( )( ) ( )( )2

2 2 2 2 24

1,2 22

a b a b a b a

x

a b

− − + ± − + − ⋅ ⋅ ⋅

⇒ = ⇒⋅ ⋅

( )2

2 2 2 2 2 24

1,2 22

a b a b a b

x

a b

+ ± + − ⋅ ⋅

⇒ = ⇒⋅ ⋅

( ) ( )2 2

2 2 2 2 2 2 2 22 4

1,2 22

a b a a b b a b

x

a b

+ ± + ⋅ ⋅ + − ⋅ ⋅

⇒ = ⇒⋅ ⋅

19

2 2 4 2 2 4 2 22 4

1,2 22

a b a a b b a bx

a b

+ ± + ⋅ ⋅ + − ⋅ ⋅⇒ = ⇒

⋅ ⋅

( )2

2 2 2 22 2 4 2 2 4

2

1,2 1,22 22 2

a b a ba b a a b b

x x

a b a b

+ ± −+ ± − ⋅ ⋅ +

⇒ = ⇒ = ⇒⋅ ⋅ ⋅ ⋅

( )( )

( )

2 2 2 2

2 2 2 21 2

21,2 2 2 2 2 22

2 22

a b a b

xa b a b

a bx

a b a b a b

x

a b

+ + −

=+ ± −

⋅ ⋅⇒ = ⇒ ⇒

⋅ ⋅ + − −

=

⋅ ⋅

2 2 2 2 2 2 2 2

1 1 12 2 22 2 2

2 2 2 2 2 2 2 2

2 2 2

2

2 2 22

2

2 2

2 2

a b a b a a a ax x x

a b a b a b

a b a b b b b bx x x

a b a b a b

b b

a a

+ + − + += = =

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⇒ ⇒ ⇒ ⇒

+ − + + + += = =

⋅ ⋅ ⋅ ⋅ ⋅

+

−

⋅

−

22

2

22

22

1 12 2 1 22.

2 1222 22

22

2

a

a a ax x x

a b b b

b xx xa

a b a

b

b

⋅ ⋅= = =

⋅ ⋅ ⋅ ⋅⇒ ⇒ ⇒

⋅ ⋅ == =

⋅ ⋅ ⋅ ⋅

Vježba 191

Riješi jednadžbu pri čemu su a i b realni brojevi: 2 2 2 2

0.a b x a x b x a⋅ ⋅ − ⋅ − ⋅ + =

Rezultat: 1

, .1 22

ax x

ab

= =


Riješi jednadžbu pri čemu su a i b realni brojevi: ( ) ( )2 2 2 20.a b x b a x a b− ⋅ − − ⋅ + ⋅ =

Rješenje 192

Ponovimo!

( ) ( ) ( ) ( ), ,2 2 2

,2

.n n n

a a a b a b a b a b a b a b a b− = − = + ⋅ − ⋅ = ⋅ ⋅ = ⋅

( ) ( )2 22 2 2 2 2

, 0 2 2, , .a a a a b a a b b a a b b a b= ≥ + = + ⋅ ⋅ + − ⋅ ⋅ + = −

,1

.n m n m

a a a a a+

= ⋅ =


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

20

( ) ( )( ) ( )

( ) ( )

2 2 2 20

2 2 2 20

2 2 2, ,

a b x b a x a b

a b x b a x a b

a a b b b a c a b

− ⋅ − − ⋅ + ⋅ =

− ⋅ − − ⋅ + ⋅ = ⇒ ⇒

= − = − − = ⋅

( ) ( )2 2 2, ,

24

1,2 2

a a b b b a c a b

b b a cx

a

= − = − − = ⋅

⇒ ⇒− ± − ⋅ ⋅

=⋅

( )( ) ( )( ) ( )

( )

222 2 2 2

4

1,2 22

b a b a a b a b

x

a b

− − − ± − − − ⋅ − ⋅ ⋅

⇒ = ⇒⋅ −

( ) ( )

( )

2 22 2 2 24

1,2 22

b a b a a b a b

x

a b

− ± − − ⋅ ⋅ ⋅ −

⇒ = ⇒⋅ −

( ) ( )

( )

2 22 2 2 24

1,2 22

b a a b a b a b

x

a b

− ± − − ⋅ ⋅ ⋅ −

⇒ = ⇒⋅ −

( ) ( )( ) ( )

( )

2 22 24

1,2 22

b a a b a b a b a bx

a b

− ± − ⋅ + − ⋅ ⋅ ⋅ −⇒ = ⇒

⋅ −

( ) ( ) ( )

( )

2 2 22 24

1,2 22

b a a b a b a b a bx

a b

− ± − ⋅ + − ⋅ ⋅ ⋅ −⇒ = ⇒

⋅ −

( ) ( ) ( )

( )

22 2 224

1,2 22

b a a ba b a ba bx

a b

− ± ⋅ + − ⋅ ⋅ ⋅=

⋅

−⇒

−

−⇒

( ) ( )( )( )

2 22 24

1,2 22

b a a b a b a b

x

a b

− ± − ⋅ + − ⋅ ⋅

⇒ = ⇒⋅ −

( ) ( )

( )

2 22 24

1,2 22

b a a b a b a bx

a b

− ± − ⋅ + − ⋅ ⋅⇒ = ⇒

⋅ −

( ) ( )

( )

22 24

1,2 22

b a a b a bx

a b

a b− ± ⋅ + − ⋅ ⋅⇒ = ⇒

⋅ −

−

( )

( )

2 2 2 22 4

1,2 22

b a a b a a b b a bx

a b

− ± − ⋅ + ⋅ ⋅ + − ⋅ ⋅⇒ = ⇒

⋅ −

21

( )

( )

( ) ( )

( )

22 22 2 2 22

1,2 1,22 22 2

b a a b a bb a a b a a b bx x

a b a b

− ± − ⋅ −− ± − ⋅ − ⋅ ⋅ +⇒ = ⇒ = ⇒

⋅ − ⋅ −

( ) ( )

( )

( )

( )

22 2 2 2

1,2 1,22 22 2

b a a b a b b a a bx x

a b a b

− ± − ⋅ − − ± −⇒ = ⇒ = ⇒

⋅ − ⋅ −

( )

( )

( ) ( ) ( )

( )

2 22 2

1,2 1,22 22 2

b a b a b a b a b ax x

a b a b

− ± − − ⋅ + ± −⇒ = ⇒ = ⇒

⋅ − ⋅ −

( ) ( ) ( )( )

( )

( ) ( ) ( )( )

( )1,2 1,22 2

2 2

b a b a b a b a b a b ax x

a b b a

− ⋅ + ± − − ⋅ + ± −⇒ = ⇒ = ⇒

⋅ − ⋅ −

( ) ( ) ( )( )

( )

( ) ( )

( )1,2 1,2 222

b a b a b a b ax x

b ab a

b a ⋅ + ± − + ±− −⇒ = ⇒ = ⇒

⋅ −⋅ −

( ) ( )

( )

( ) ( )

( )

( )

( )

( )

( )

1 1 12 2 2

2 22 2 22

b a b a b a b a b bx x x

b a b a b a

b a b a a ab a b ax xx

b a b

a a

b b

ab a

+ + − + + − += = =

⋅ − ⋅ − ⋅ −⇒ ⇒ ⇒ ⇒

+ − + + ++ − −= ==

⋅ − ⋅⋅

+

−−

−

−

( )

( )

( )

( )

2

1 1 1

2

2.

2

2 2 22 2

2

2

b b bx x x

b a b a b a

a a ax x x

b a b a b a

⋅ ⋅= = =

⋅ − ⋅ − −⇒ ⇒ ⇒

⋅ ⋅= = =

⋅ − ⋅ − −

Vježba 192

Riješi jednadžbu pri čemu su a i b realni brojevi: ( ) ( )2 2 2 20.a b x a b x a b− ⋅ + − ⋅ + ⋅ =

Rezultat: , .1 2

b ax x

b a b a= =

− −


Riješi jednadžbu pri čemu su a i b realni brojevi: ( )2 2 22 0.x a x a b− ⋅ ⋅ + − =

Rješenje 193

Ponovimo!

( ) ( ), ,2

, .2 2

, 0n n n

a a a b a b a b a b a a a− = ⋅ = ⋅ ⋅ = ⋅ = ≥


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

22

( ) ( )2 2 22 02 2 2

2 02 2

1 , 2 ,

x a x a bx a x a b

a b a c a b

− ⋅ ⋅ + − =− ⋅ ⋅ + − = ⇒ ⇒

= = − ⋅ = −

2 21 , 2 ,

24

1,2 2

a b a c a b

b b a cx

a

= = − ⋅ = −

⇒ ⇒− ± − ⋅ ⋅

=⋅

( ) ( ) ( )2 2 22 2 4 1

1,2 2 1

a a a b

x

− − ⋅ ± − ⋅ − ⋅ ⋅ −

⇒ = ⇒⋅

2 24

2 2 2 22 4 4 4 2 4

1,2 1 22 2

4

,

a a a bx x

aa ba⋅ ± ⋅ − ⋅ + ⋅ ⋅ ± + ⋅⇒ = ⇒

⋅ − ⋅= ⇒

2 22 4 2 4 2 2

1,2 1,2 1,22 2 2

a b a b a bx x x

⋅ ± ⋅ ⋅ ± ⋅ ⋅ ± ⋅⇒ = ⇒ = ⇒ = ⇒

( ) ( )2

1,2 1,2 1,2

22

22

2

2

a b a ba bx x x

⋅ ± ⋅ ±⋅ ± ⋅⇒ = ⇒ = ⇒ = ⇒

1.

1,22

x a bx a b

x a b

= +⇒ = ± ⇒

= −

Vježba 193

Riješi jednadžbu pri čemu su a i b realni brojevi: ( )2 2 22 0.x a x b a− ⋅ ⋅ − − =

Rezultat: , .1 2

x a b x a b= + = −


Zadana je kvadratna jednadžba i jedno njezino rješenje x1. Izračunaj vrijednost parametra m.

( )2

1 2 6 0 , 3.1

m x x m x+ ⋅ + ⋅ − ⋅ = =

Rješenje 194

Ponovimo!

Parametar

Vladimir Anić, Ivo Goldstein, Rječnik stranih riječi, Novi Liber, Zagreb, 2002.

Veličina, obično realna varijabla, čije vrijednosti služe za razlikovanje elemenata nekog skupa točaka

funkcija, jednadžbi ili drugih matematičkih objekata.

Bratoljub Klaić, Rječnik stranih riječi, Nakladni zavod MH, Zagreb, 1983.

Veličina o kojoj ovisi funkcija ili oblik krivulje.


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

( ) [ ] ( )32 2

1 2 6 0 1 3 2 3 6 0m x x mxm m+ ⋅ + ⋅ − ⋅ = ⇒ ⇒ + ⋅ + ⋅ − ⋅ == ⇒

( )1 9 6 6 0 9 9 6 6 0 9 6 9 6 3 15m m m m m m m⇒ + ⋅ + − ⋅ = ⇒ ⋅ + + − ⋅ = ⇒ ⋅ − ⋅ = − − ⇒ ⋅ = − ⇒

/ :3 315 5.m m⇒ ⋅ = − ⇒ = −

23

Vježba 194

Zadana je kvadratna jednadžba i jedno njezino rješenje x1. Izračunaj vrijednost parametra m.

( )2

1 2 6 0 , 2.1

m x x m x+ ⋅ + ⋅ − ⋅ = =

Rezultat: m = 4.


U ovisnosti o parametru m opiši vrstu rješenja kvadratne jednadžbe ( )2

1 3 2 0.m x x+ ⋅ + ⋅ − =

Rješenje 195

Ponovimo!

, 0 , ., 0a b a b

a b c a b cc c c c

> > ⇒ > < > ⇒ <

Parametar

Vladimir Anić, Ivo Goldstein, Rječnik stranih riječi, Novi Liber, Zagreb, 2002.

Veličina, obično realna varijabla, čije vrijednosti služe za razlikovanje elemenata nekog skupa točaka

funkcija, jednadžbi ili drugih matematičkih objekata.

Bratoljub Klaić, Rječnik stranih riječi, Nakladni zavod MH, Zagreb, 1983.

Veličina o kojoj ovisi funkcija ili oblik krivulje.


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


20 , 0a x b x c a⋅ + ⋅ + = ≠

je broj

24 .D b a c= − ⋅ ⋅

• Ako je D > 0, jednadžba ima dva realna i različita rješenja.



Uočimo da mora biti

( )2

1 3 2 0 1 .10m x x m m+ ⋅ + ⋅ − = ≠ ⇒ ≠ −⇒ +

Dalje imamo:

( )( )

2 1 , 3 , 21 3 2 021 3 2 0

241 , 3 , 2

a m b cm x xm x x

D b a ca m b c

= + = = −+ ⋅ + ⋅ − =+ ⋅ + ⋅ − = ⇒ ⇒ ⇒

= − ⋅ ⋅= + = = −

( ) ( ) ( )2

3 4 1 2 9 8 1 9 8 8 8 17.D m D m D m D m⇒ = − ⋅ + ⋅ − ⇒ = + ⋅ + ⇒ = + ⋅ + ⇒ = ⋅ +

Opisi:

• 17

8 17 0 8 17 80 /7 ,81 :8

m m m mD ⇒ ⋅ + > ⇒ ⋅ > − ⇒ ⋅ > > −> − ⇒

jednadžba ima dva realna i različita rješenja.

• 17

8 17 0 8 17 80 /7 ,81 :8

m m m mD ⇒ ⋅ + = ⇒ ⋅ = − ⇒ ⋅ = = −= − ⇒

jednadžba ima jedno dvostruko realno rješenje.

• 17

8 17 0 8 17 80 /7 ,81 :8

m m m mD ⇒ ⋅ + < ⇒ ⋅ < − ⇒ ⋅ < < −< − ⇒

jednadžba ima kompleksno – konjugirana rješenja.

24

Vježba 195

U ovisnosti o parametru m opiši vrstu rješenja kvadratne jednadžbe

( )2

1 2 2 0.m x m x m− ⋅ + ⋅ ⋅ + + =

Rezultat: 1m ≠

2, jednadžba ima dva realna i različita rješenja,m <

2, jednadžba ima jedno dvostruko realno rješenje,m =

2, jednadžba ima kompleksno konjugirana rješenja.m > −


Odredi jednadžbe tangenata na parabolu 2

2 2 2y x x= ⋅ + ⋅ + koje prolaze točkom A(1, – 12).

Rješenje 196

Ponovimo!

( )2 2 2

2 .a b a a b b− = − ⋅ ⋅ +


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅ Jednadžba pravca oblika

y k x l= ⋅ +

naziva se eksplicitni oblik jednadžbe pravca ili kraće, eksplicitna jednadžba pravca. Broj k naziva se

koeficijent smjera pravca. Broj l nazivamo odsječak pravca na osi y.


20 , 0a x b x c a⋅ + ⋅ + = ≠

je broj

24 .D b a c= − ⋅ ⋅




Budući da točka A pripada tangenti (pravcu), uvrstit ćemo njezine koordinate u jednadžbu pravca

(tangente) y = k · x + l.

( ) ( ), 1, 1212 1 12 12 12 .

A x y Ak l k l k l l k

y k x l

= −⇒ − = ⋅ + ⇒ − = + ⇒ + = − ⇒ = − −

= ⋅ +

Jednadžba tangente ima oblik

1212 .

l ky k x k

y k x l

= − −⇒ = ⋅ − −

= ⋅ +

Da bismo odredili presjek pravca (tangente) i parabole moramo riješiti sustav jednadžba.

metoda

komparacije

222 2 2

2 2 2 1212

y x xx x k x k

y k x k

= ⋅ + ⋅ +⇒ ⇒ ⋅ + ⋅ + = ⋅ − − ⇒

= ⋅ − −

( )2 2

2 2 2 12 0 2 2 14 0.x x k x k x k x k⇒ ⋅ + ⋅ + − ⋅ + + = ⇒ ⋅ + − ⋅ + + =

Ova jednadžba ima jedno rješenje (dvostruko realno rješenje), ako je D = 0.

25

Zato slijedi:

( )( )

22 2 14 02

2 2 14 02 , 2 ,

2014

0

4

x k x kx k x k

a b k c k

D

b a c

⋅ + − ⋅ + + =⋅ + − ⋅ + + = ⇒ ⇒ ⇒

= =

>

>− − ⋅+ ⋅=

( ) ( ) ( ) ( )2 2

2 4 2 14 0 2 8 14 0k k k k⇒ − − ⋅ ⋅ + = ⇒ − − ⋅ + = ⇒

22 2 12 108 0

4 4 112 8 0 12 108 01 , 12 , 108

k kk k k k k

a b c

− ⋅ − =⇒ − ⋅ + − − ⋅ = ⇒ − ⋅ − = ⇒ ⇒

= = − = −

( ) ( ) ( )1 , 12 , 108 2

12 12 4 1 1082

4 1,2 2 11,2 2

a b c

kb b a c

ka

= = − = −− − ± − − ⋅ ⋅ −

⇒ ⇒ = ⇒− ± − ⋅ ⋅ ⋅=

⋅

12 24

112 144 432 12 576 12 24 21,2 1,2 1,2 12 242 2 2

2 2

k

k k k

k

+=

± + ± ±⇒ = ⇒ = ⇒ = ⇒ ⇒

−=

36181 12 1

.12 6

22 2

36

2

12

22

k k k

kk k

= = =⇒ ⇒ ⇒

= −= − = −

Računamo odsječak l.

• 12

12 18 301 118

l kl l

k

= − −⇒ = − − ⇒ = −

=

• ( )12

12 6 12 6 6.2 2 26

l kl l l

k

= − −⇒ = − − − ⇒ = − + ⇒ = −

= −

Jednadžbe tangenata glase:

• 18

3018 30y k x l y x

k

l= ⋅ + ⇒ ⇒ = ⋅

=

−−

=

• 6 6.6

6

ky k x l y x

l

= −

== ⋅ + ⇒ ⇒ = − ⋅

−−

Vježba 196

Odredi koeficijente smjerova tangenata na parabolu 2

2 2 2y x x= ⋅ + ⋅ + koje prolaze točkom

A(1, 11).

Rezultat: 11, 5.1 2

k k= = −


Ako su x1 i x2 rješenja jednadžbe2

0x p x q+ ⋅ + = izračunaj ( )2

.1 2

x x−

+

Rješenje 197

Ponovimo!

,1

.1

,

n na a n n

n an nb b a

−= = =

26

Jednadžba oblika

0,2

a x b x c⋅ + ⋅ + =

(a ≠ 0, b i c su realni brojevi) naziva se kvadratna jednadžba. Svaki broj x (realan ili kompleksan) koji

zadovoljava tu jednadžbu naziva se rješenje kvadratne jednadžbe.

Rješenja x1 i x2 kvadratne jednadžbe

20a x b x c⋅ + ⋅ + =

zadovoljavaju Vièteove formule:

,2

.1 2 1

b cx x x x

a a+ = − ⋅ =

1 , ,22 0

01 2 11 , , 1 2

a b p c qpx p x q

x p x q x xbx xa b p c q

a

= = =+ ⋅ + =

+ ⋅ + = ⇒ ⇒ ⇒ + = − ⇒+ = −= = =

.1 2

x x p⇒ + = −

Sada je:

( )( ) ( )

2 1 1 1.

1 2 2 2 2

1 2

x x

ppx x

−+ = = =

−+

Vježba 197

Ako su x1 i x2 rješenja jednadžbe2

0x p x q+ ⋅ + = izračunaj ( )1

.1 2

x x−

+

Rezultat: 1

.p

−


Znajući da su m i n rješenja jednadžbe 2

0x x c+ − = pojednostavnite izraz

3 3 2 25 5 .I m n m n m n= + − ⋅ ⋅ − ⋅ ⋅

Rješenje 198

Ponovimo!

( ) ( ) ( )23 3 2 2 2

, ,2

1.2

na b a b a a b b a a b b a b n+ = + ⋅ − ⋅ + + ⋅ ⋅ + = + =

,1

.:n m n m

a a a a a−

= =


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +

Jednadžba oblika

0,2

a x b x c⋅ + ⋅ + =

(a ≠ 0, b i c su realni brojevi) naziva se kvadratna jednadžba. Svaki broj x (realan ili kompleksan) koji

zadovoljava tu jednadžbu naziva se rješenje kvadratne jednadžbe.

Rješenja x1 i x2 kvadratne jednadžbe

20a x b x c⋅ + ⋅ + =

zadovoljavaju Vièteove formule:

,2

.1 2 1

b cx x x x

a a+ = − ⋅ =

3 3 2 25 5I m n m n m n= + − ⋅ ⋅ − ⋅ ⋅ ⇒

27

( ) ( ) ( )2 2

5I m n m m n n m n m n⇒ = + ⋅ − ⋅ + − ⋅ ⋅ ⋅ + ⇒

( ) ( ) ( )2 2

2 3 5I m n m m n n m n m n m n⇒ = + ⋅ + ⋅ ⋅ + − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⇒

( ) ( )( ) ( )2 2

2 3 5I m n m m n n m n m n m n⇒ = + ⋅ + ⋅ ⋅ + − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⇒

( ) ( )( ) ( )2

3 5I m n m n m

bm n

an m n

ma

m nc

n

⇒ = + ⋅ + − ⋅ ⋅ − ⋅ ⋅ ⋅ + ⇒ ⇒

+ = −

⋅ =

220

1 , 13

,5

b b c c bI

a a a a

x c

a b c ca

x⇒ = − ⋅ − − ⋅ − ⋅ ⋅ −

+ − =

=⇒

−⇒

= =

( ) ( ) ( )( ) ( ) ( )2

1 1 1 23 5 1 1 3 5 1

1 1 1 1 1

c cI I c c

− −⇒ = − ⋅ − − ⋅ − ⋅ ⋅ − ⇒ = − ⋅ − − ⋅ − − ⋅ − ⋅ − ⇒

( ) ( )1 1 3 5 1 3 5 1 8 .I c c I c c I c⇒ = − ⋅ + ⋅ − ⋅ ⇒ = − − ⋅ − ⋅ ⇒ = − − ⋅

Vježba 198

Znajući da su m i n rješenja jednadžbe 2

1 0x x+ − = pojednostavnite izraz

3 3 2 25 5 .I m n m n m n= + − ⋅ ⋅ − ⋅ ⋅

Rezultat: 7.

Zadatak 199 (Katarina, srednja škola)

Zadana je kvadratna jednadžba ( ) ( )1 2 2p x x x p x⋅ ⋅ + + = ⋅ ⋅ + pri čemu je x nepoznanica, a p

neki realni broj. Za koje p jednadžba ima realna rješenja?

Rješenje 199

Ponovimo!

( )21 2 2 2

2 0, , ,, .n m n m

a a a a a a a b b a b a a R+

= ⋅ = − ⋅ ⋅ + = − ≥ ∈


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


20 , 0a x b x c a⋅ + ⋅ + = ≠

je broj

24 .D b a c= − ⋅ ⋅




Preoblikujemo jednadžbu tako da je zapišemo u općem obliku:

( ) ( )2 2

1 2 2 2 2 2p x x x p x p x p x x p x⋅ ⋅ + + = ⋅ ⋅ + ⇒ ⋅ + ⋅ + = ⋅ ⋅ + ⋅ ⇒

2 2 2 22 2 2 0 2 2 0p x p x x p x p x x p x⇒ ⋅ + ⋅ + − ⋅ ⋅ − ⋅ = ⇒ ⋅ − ⋅ − ⋅ + = ⇒

28

( ) ( )2 2 2

2 2 0 2 2 0p x x p x p x p x⇒ ⋅ − ⋅ − ⋅ + = ⇒ − ⋅ − ⋅ + = ⇒

( )2 2 , , 22 2 0

242 , , 2

a p b p cp x p x

D b a ca p b p c

= − = − =− ⋅ − ⋅ + =⇒ ⇒ ⇒

= − ⋅ ⋅= − = − =

( ) ( ) ( )2 2 2

4 2 2 8 2 8 16D p p D p p D p p⇒ = − − ⋅ − ⋅ ⇒ = − ⋅ − ⇒ = − ⋅ + ⇒

( )2

4 .0D p⇒ = − ≥

Budući da je D ≥ 0, zaključujemo da jednadžba uvijek ima realna rješenja za svaki realni broj p.

Vježba 199

Zadana je kvadratna jednadžba ( ) ( )1 2 2p x x x p x⋅ ⋅ + = ⋅ ⋅ + − pri čemu je x nepoznanica, a p

neki realni broj. Za koje p jednadžba ima realna rješenja?

Rezultat: .p R∈

Zadatak 200 (Sonja, maturantica)

Riješite jednadžbu: 3 2

12 13 13 12 0.x x x⋅ − ⋅ − ⋅ + =

Rješenje 200

Ponovimo!

( ) ( ),1 3 2 2

, .3

:n m n m

a a a a a a b a b a a b b−

= = + = + ⋅ − ⋅ +


( ) ( ), .a b c a b a c a b a c a b c⋅ + = ⋅ + ⋅ ⋅ + ⋅ = ⋅ +


0 0 ili 0 il .i 0a b a b a b⋅ = ⇔ = = = =


jedinice

, 0 ., 1a n a

n nb n b

⋅= ≠ ≠

⋅

Kubna jednadžba ima oblik

3 2.0 , 0a x b x c x d a⋅ + ⋅ + ⋅ + = ≠

Kubna jednadžba općenito ima tri rješenja.

( ) ( )3 2 3 212 13 13 12 0 12 12 13 13 0x x x x x x⋅ − ⋅ − ⋅ + = ⇒ ⋅ + + − ⋅ − ⋅ = ⇒

( ) ( ) ( ) ( ) ( )3 2

12 1 13 1 0 12 1 1 13 1 0x x x x x x x x⇒ ⋅ + − ⋅ ⋅ + = ⇒ ⋅ + ⋅ − + − ⋅ ⋅ + = ⇒

( ) ( )( ) ( ) ( )2 21 12 1 13 0 1 12 12 12 13 0x x x x x x x x⇒ + ⋅ ⋅ − + − ⋅ = ⇒ + ⋅ ⋅ − ⋅ + − ⋅ = ⇒

( ) ( )1 02

1 12 25 12 0 .2

12 25 12 0

xx x x

x x

+ =⇒ + ⋅ ⋅ − ⋅ + = ⇒

⋅ − ⋅ + =

Riješimo prvu jednadžbu (linearnu jednadžbu).

1 0 1.1

x x+ = ⇒ = −

Riješimo drugu jednadžbu (kvadratnu jednadžbu).

29

12 , 25 , 122

2 12 25 12 0212 25 12 0

412 , 25 , 12

2,3 2

a b c

x xx x

b b a ca b c x

a

= = − =

⋅ − ⋅ + =⋅ − ⋅ + = ⇒ ⇒ ⇒

− ± − ⋅ ⋅= = − = =

⋅

( ) ( )2

25 25 4 12 12 25 625 576 25 49

2,3 2,3 2,32 12 24 24x x x

− − ± − − ⋅ ⋅ ± − ±⇒ = ⇒ = ⇒ = ⇒

⋅

425 7 32

22 2 225 7 324 24.

2,3 25 7 18

3

324

3

2

24

18

23 3 324 4 442

xx x x

x

x x x x

+== = =

±⇒ = ⇒ ⇒ ⇒ ⇒

−= = = =

Vježba 200

Riješite jednadžbu: 3 2

6 7 7 6 0.x x x⋅ − ⋅ − ⋅ + =

Rezultat: 2 3

1 , , .1 2 33 2

x x x= − = =

4 ponovimo! za realni broj x njegova je apsolutna vrijednost (modul) broj x koji odre đujemo na...

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