Másodfokú egyenletre visszavezethető magasabb fokú egyenletek
EMLÉKEZTETŐ
| új változó bevezetésének módszere: |
Vannak olyan egyenlettípusok, amelyeket a másodfokú egyenletre vonatkozó ismereteink segítségével
meg tudunk oldani. Például tekintsünk olyan negyedfokú egyenletet, amelyben az ismeretlen csak negyed- és másodfokon fordul elő: x4 -5x² + 6 = 0. Vezessük be új ismeretlennek az y = x²-et. Ekkor y² -5y + 6 = 0 Ennek a gyökei y1 = 3, y2 = 2. Ebből x1 = √3, x2 = -√3, x3 = √2 és x4 = -√2 négy valós gyök adódik. Vagy például az x -23√y + 132 = 0 egyenlet az y = √x helyettesítéssel y² -23y +132 = 0 másodfokú egyenletté alakul. |
FELADATOK
15.1. Oldd meg a valós számok halmazán a következő egyenleteket!
a) x4 -5x² +4 = 0
y² -5y +4 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
b) x4 -8x² -9 = 0
y² -8y -9 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
c) x4 - 4,25x² +1 = 0
y² -4,25y +1 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
d) x4 +29x² + 100 = 0
y² +29y +100 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
a) x4 -5x² +4 = 0
y² -5y +4 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
b) x4 -8x² -9 = 0
y² -8y -9 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
c) x4 - 4,25x² +1 = 0
y² -4,25y +1 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
d) x4 +29x² + 100 = 0
y² +29y +100 = 0
y1 = = x²
x1 =
x2 =
y2 = = x²
x3 =
x4 =
| Max p. | Kapott p. |
| 24 pont |
15.2. Oldd meg a valós számok halmazán a következő egyenleteket!
a) x6 -26x³ -27 = 0
y² -26y -27 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
b) x6 +126x³ + 125 = 0
y² +126y +125 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
c) x6 -1 = 0
y² -1 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
d) x8 -17x4 +16 = 0
y² -17y 16 = 0
y1 = = x4
x1 =
x2 =
y2 = = x4
x3 =
x4 =
a) x6 -26x³ -27 = 0
y² -26y -27 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
b) x6 +126x³ + 125 = 0
y² +126y +125 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
c) x6 -1 = 0
y² -1 = 0
y1 = = x³
x1 =
y2 = = x³
x2 =
d) x8 -17x4 +16 = 0
y² -17y 16 = 0
y1 = = x4
x1 =
x2 =
y2 = = x4
x3 =
x4 =
| Max p. | Kapott p. |
| 18 pont |
15.3. Oldd meg a valós számok halmazán a következő egyenleteket!
a) `x-3sqrt(x)-4=0`
y² -3y -4 = 0
y1 = = √x
x1 =
y2 = = √x
x2 =
b) `x-23sqrt(x)+132=0`
y² -23y +132 = 0
y1 = = √x
x1 =
y2 = = √x
x2 =
c) `root(3)(x²)-root(3)(x)-30=0`
y² -y -30 = 0
y1 = = `root(3)(x)`
x1 =
y2 = = `root(3)(x)`
x2 =
d) `sqrt(x)+2root(4)(x)+4=0`
y² +2y +4 = 0
y1 = = `root(4)(x)`
x1 =
y2 = = `root(4)(x)`
x2 =
a) `x-3sqrt(x)-4=0`
y² -3y -4 = 0
y1 = = √x
x1 =
y2 = = √x
x2 =
b) `x-23sqrt(x)+132=0`
y² -23y +132 = 0
y1 = = √x
x1 =
y2 = = √x
x2 =
c) `root(3)(x²)-root(3)(x)-30=0`
y² -y -30 = 0
y1 = = `root(3)(x)`
x1 =
y2 = = `root(3)(x)`
x2 =
d) `sqrt(x)+2root(4)(x)+4=0`
y² +2y +4 = 0
y1 = = `root(4)(x)`
x1 =
y2 = = `root(4)(x)`
x2 =
| Max p. | Kapott p. |
| 16 pont |
15.4. Oldd meg a valós számok halmazán a következő egyenleteket!
a) (x +1)4 -13(x +1)² + 36 = 0
y² -13y +36 = 0
y1 = = (x +1)²
x1 =
x2 =
y2 = = (x +1)²
x3 =
x4 =
b) (x -1/2)4 + (x -1/2)² = 0
y² +y = 0
y1 = = (x -1/2)²
x1 =
x2 =
y2 = = (x -1/2)²
x3 =
x4 =
a) (x +1)4 -13(x +1)² + 36 = 0
y² -13y +36 = 0
y1 = = (x +1)²
x1 =
x2 =
y2 = = (x +1)²
x3 =
x4 =
b) (x -1/2)4 + (x -1/2)² = 0
y² +y = 0
y1 = = (x -1/2)²
x1 =
x2 =
y2 = = (x -1/2)²
x3 =
x4 =
| Max p. | Kapott p. |
| 12 pont |
15.5. Szorzattá alakítás segítségével oldd meg a következő egyenleteket!
a) x³ -9x = 0
x() = 0
x1 =
x2 =
x3 =
b) x³ -5x = 0
x() = 0
x1 =
x2 =
x3 =
c) x³ + 5x² + 4x = 0
x() = 0
x1 =
x2 =
x3 =
d) (x +1)³ - 0,8(x +1)² + 0,24(x + 1) = 0
x +1 = y
(x +1)() = 0
x1 =
x2 =
x3 =
a) x³ -9x = 0
x() = 0
x1 =
x2 =
x3 =
b) x³ -5x = 0
x() = 0
x1 =
x2 =
x3 =
c) x³ + 5x² + 4x = 0
x() = 0
x1 =
x2 =
x3 =
d) (x +1)³ - 0,8(x +1)² + 0,24(x + 1) = 0
x +1 = y
(x +1)() = 0
x1 =
x2 =
x3 =
| Max p. | Kapott p. |
| 16 pont |
1. feladatsor
NÉV:EREDMÉNY:
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