Question

Assalamu'alaikum... Tolong bantu jawab dong... Aku ga ngeri sama yg nomor 3
Soal UM UGM 2016

Ajarkan langkah²nya... Supaya ana ngerti

Answer 1

3. 3x²+x-3 = 0 => akar : α dan β

α+β = -b/a

      = -1/3

αβ = c/a

     = (-3)/3 = -1

Akar persamaan yang baru :

c = 2+(1/(α+1)) , d = 2+(1/(β+1))

rumus PK baru :

x² - (c+d)x + c.d = 0

c+d = 2+(1/(α+1)) + 2+(1/(β+1))

     = 4 + ( (β+1)+(α+1) )/( (α+1)(β+1) )

     = 4 + (2+(α+β) )/( αβ+1+(α+β) ) <= α+β = -1/3 , αβ = -1

    = 4 + (2 - 1/3 )/( -1+1 - 1/3 )

    = 4 + (5/3)/(-1/3)

   = 4 - 5

c+d = -1

cd = (2+(1/(α+1)) ).(2+(1/(β+1)) )

     = 4 + 2.(1/(β+1)) + 2.(1/(α+1)) + (1/(α+1)).(1/(β+1))

     = 4 + 2.( (1/(β+1)) + (1/(α+1)) ) + 1/( (α+1).(β+1) )

     = 4 + 2.( ( (α+1)+(β+1) )/( (α+1).(β+1) ) ) + 1/( (α+1).(1/(β+1) ) <= (α+1).(β+1) = -1/3 , (α+1)+(β+1) = 5/3

    = 4 + 2.(5/3)/(-1/3) + 1/(-1/3)

    = 4 + 2.(-5) - 3

    = 4 - 10 - 3

c.d = -11

Persamaannya adalah :

x² - (-1)x + (-11) = 0

x² + x - 11 = 0 (A)

4. x² - (3p-2)x + 2p+8 = 0  => akarnya x₁ dan x₂

x₁ + x₂ = - (-(3p-2))/1

x₁ + x₂ = 3p - 2

x₁.x₂ = (2p+8)/1 = 2p+8

x₁,p,x₂ membentuk barisan geometri

r = p/x₁ = x₂/p

    p² = x₁.x₂

     p² = 2p+8

     p²-2p-8 = 0

  (p-4)(p+2) = 0

p = 4 atau -2

di soal dibilang bahwa p positif, maka :

p = 4

x₁ + x₂ = 3p - 2 <= p = 4

x₁ + x₂ = 3.4 - 2

x₁+x₂ = 10

x₁+x₂+p = 10 + 4 = 14 (E)

5. x²-25x+c = 0 => akar : a dan b (keduanya prima , b > a)

a+b = -(-25) = 25

ab = c

bilangan prima : 2,3,5,7,11,13,17,19,23,29,31,33, dan seterusnya

diantara bilangan diatas yang kalau ditambah menghasilkan 25 adalah 2+23 , berarti a dan b adalah 2 dan 23

a = 2 , b = 23 (karena b > a)

berarti ab = c = 2.23 = 46

                      c = 46

3a-b+c = 3.2 - 23 + 46

           = 6 + 23

           = 29 (C)