Question

Wyznacz trójmian kwadratowy o pierwiastkach x1 i x2 i podanym zbiorze wartości Y. a)x1=-1 x2=3 Y=<-2,∞) b)x1=-4 x2 =0 Y= (-∞,¼> c)x1=-8 x2=2 Y=(-∞,3> d) x1=0 x2=10 Y=<-1,∞). Rozwiąż nierówność √6x²+√3x>0 . Wyznacz liczbe wyrazów skonczonego ciągu arytmetycznego oraz oblicz ich sume 8,5,2,...,-25 . z góry dzieki

Answer 1

z.1

a) x1 = -1 ; x2 = 3

Y = < -2; + ∞ )

Mamy

y = a(x -x1)(x -x2)

y = a(x +1)(x -3) = a(x² -3x + x - 3) = a(x² - 2x - 3) = ax² -2a x - 3a

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

q = f(p) = f(1) = a*1² -2*1 -3*1 = - 4a

oraz q = -2

zatem -4a = -2  --> a = -2 : (-4) = 1/2

mamy więc

y = (1/2)x² -2(1/2) x - 3(1/2) = (1/2) x² - x - 3/2

Odp. y = 0,5 x² - x - 1,5

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b)

x1 = -4; x2 = 0

Y = ( -∞ ; 1/4>

y =a(x +4)(x - 0) = a x ( x +4) = ax² + 4ax

p = [x1 +x2]/2 = [-4 + 0]/2 = -4/2 = -2

q = f(p) = f( -2) = a (-2)² + 4a(-2) = 4a - 8a = - 4a

ale q = 1/4 zatem

-4a = 1/4

a = -1/16

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

Odp. y = (-1/16) x² - (1/4) x

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c)

x1 = - 8

x2 = 2

Y = ( -∞ ; 3 > = ( - ∞ ; q >

y = a (x +8)(x -2) = a(x²  -2x +8x - 16) = ax² +6ax - 16a

p = [-8 + 2]/2 = -6/2 = -3

q = f(p) = f(-3) = a(-3)² + 6a(-3) - 16a = 9a - 18a - 16a = - 25 a

ale q = 3

zatem -25 a = 3  --> a = -3/25

czyli  y = (-3/25)x² + 6(-3/25)x - 16(-3/25) = (- 3/25) x² - (18/25)x + (48/25)

Odp. y = -0,12 x² = 0,72x + 1,92

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d)

x1 = 0; x2 = 10

Y = < -1 ; + ∞ )

q = - 1

y = a(x -0)(x -10) = ax(x -10) = ax² - 10a x

p = [x1 + x2]/2 = [0 + 10]/2 = 10/2 = 5

q = f(p) = f(5) = 25a - 50a = - 25a

zatem   -25 a = - 1

a = 1/25 = 0,04

czyli

y = 0,04 x² -0,4 x

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z.2

√6 x² + √3 x > 0

√6 x ( x + 1/√2) >  0

x1 = 0  oraz  x2 = -1/√2

a = √6 > 0  zatem

x ∈ ( -∞ ; -1/√2) u ( 0 ; + ∞ )

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z.3

8, 5,2, -1,-4, ... , -25  - ciąg arytmetyczny

zatem

a1 = 8

r = 5 - 8 = -1 -2 = - 3

an = -25

an = a1 +(n -1)r = 8 +(n -1)(-3) = 8 +3 - 3n = 11 - 3n

an = 11 - 3n

zatem  11 - 3n = -25

3n = 11 + 25 = 36 / : 3

n = 12

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S12 =0,5* [a1 + a12]*12 = 6*[8  +(-25)] = 6* (- 17) = - 102

S12 = - 102

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Odp. Ten ciąg ma 12 wyrazów. Suma tych wyrazów jest rowna

S12 = - 102

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