Question

Wyznacz trójmian kwadratowy o pierwiastkach x1,x2 i podanym zbiorze wartosci Y. a)x1=-4 x2=3 y=<-2,∞) b)x1=-4 x2=0 Y=(-∞,¼> . Rozwiąż równanie x²=(1-x)(1+x). Rozłóż wielomian w na czynniki; w(x)=-12x⁵+12x³-3x.Rozłóż wielomian w na czynniki, grupujac jego wyrazy w(x)½x³-⅙x²-3x+1, w(x)=x⁵+x⁴-2x³-2x²+x+1. oblicz log⅕ 5√5

Answer 1

z.1

a) x1 = -4 oraz x2 = 3    oraz     Y = < -2; +∞ )

Mamy

p = [x1+ x2]/2 = [-4 + 3]/2 = -1/2

q = -2

Mamy zatem

y = a(x -x1)(x-x2) = a(x - p)² + q

czyli

a( x +4)( x -3) = a[x +1/2]² - 2

a[x² +x -12] = a[x² +x + 1/4) - 2

a x² + a x - 12 a = a x² = ax + (1/4)a -2

-12 a -(1/4)a = -2

-12,25 a = -2

a = -2/(-12,25) = 8/49

zatem

y = (8/49)(x +4)(x -3)  lub   y = (8/49)x² +(8/49)x - 96/49

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b)

x1 = -4 oraz    x2 = 0  oraz  Y = ( -∞ ; 1/4>

Mamy

p = [x1 + x2]/2 = -4/2 = -2

q = 1/4

zatem

a(x-0)(x +4) = a[x +2]² + 1/4

ax(x +4) = a[x² +4x = 4] + 1/4

ax² + 4a x = a x² + 4a x + 4a + 1/4

0 = 4a + 1/4

4a = - 1/4  / : 4

a = - 1/16

mamy więc

y = (-1/16)x(x +4)   lub  y = (-1/16) x² - (1/4) x

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z.2

x² = (1 -x)( 1 +x)

x² = 1 - x²

2x² = 1

x² = 1/2

x = - 1/√2  lub  x = 1/√2  lub inaczej x = - √2/2  lub  x = √2/2

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W(x) = - 12 x⁵ + 12 x³ -3 x = -3x (4 x⁴ - 4x² + 1) = -3x ( 2x² - 1)² =

= -3x[ (√2 x -1)(√2 x + 1)]² = -3x (√3 -1)² * (√3 x + 1)²

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W(x) = x⁵ + x⁴ -2 x³ - 2 x² + x + 1 = x(x⁴ - 2x² + 1) + (x⁴ -2 x² + 1 ) =

= (x +1) (x² -1 )² = (x+1) (x-1)²(x+1)² = (x +1)³(x -1)²

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W(x) = (1/2) x³ - (1/6) x² - 3x +1 = (1/2) x[x² -6] -(1/6)[x² - 6] =

= (0,5x - 1/6) (x² - 6) = (0,5 x - 1/6)( x -√6)(x + √6)

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log ¹/₅ 5√5 = log ¹/₅ 5^(3/2) = (3/2) log ¹/₅ 5 = (3/2) *(-1) = (-3/2) = -1,5

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