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PK baru yg memiliki 2x₁+5 dan 2x₂+5:
x² - (2x₁ + 5 + 2x₂ + 5)x + (2x₁+5)(2x₂+5) = 0
x² - (2(x₁+x₂)+10)x + 4(x₁.x₂)+10(x₁+x₂)+25 = 0
x² - (2(4)+10)x + 4(3)+10(4)+25 = 0
x² - 18x + 77 = 0............opsi D
3. x² - 5x + 6 = 0 ; mempunyai akar2 x₁ dan x₂
PK baru yang memiliki akar2 x₁-3 dan x₂-3
x² - (x₁ -3 + x₂ -3)x + (x₁-3)(x₂-3) = 0
x² - ((x₁+x₂)-6)x + x₁x₂ -3(x₁+x₂)+9 = 0
x² - (5-6)x + 6-3(5)+9 = 0
x² + x = 0......................opsi C