a.x1¹+x2²
b.(x1–x2)²
c.1/x1+1/x2
d.x1/x2+x2/x1
Jika x¹ dan x² akar akar persamaan kuadrat x²–2x+4=0 tentukan nilai berikut:
Jawaban:
Persamaan Kuadrat
Penjelasan dengan langkah-langkah:
x²-2x+4=0
a. x₁²+x₂²= (x₁+x₂)²-2x₁.x₂
=(-b/a)²-2(c/a)
=(-(-2)/1)²-2(4/1)
=(2)²-8
=4-8
=-4
b. (x₁-x₂)²=(x₁)²-2x₁.x₂+(x₂)²
=x₁²+x₂²-2x₁.x₂
=(x₁+x₂)²-2x₁.x₂-2x₁.x₂
=(x₁+x₂)²-4(x₁.x₂)
=(-b/a)²-4(c/a)
=(-(-2)/1)²-4(4/1)
=(2)²-16
=4-16
=-12
c. 1/x₁ + 1/x₂ = x₂/( x₁.x₂ ) + x₁/ ( x₁.x₂ )
= (x₁ + x₂) / (x₁.x₂)
= ( -b/a ) / (c/a)
= (-(-2)/1) / (4/1)
= 2/4
= 1/2
d. x₁/x₂ + x₂/x₁ = x₁² / (x₂.x₁ )+ x₂² / (x₁.x₂)
= (x₁² + x₂²) / (x₁.x₂)
= [(x₁+x₂)²-2(x₁.x₂)] / (x₁.x₂)
= [(-b/a)² -2(c/a)] / (c/a)
= [(-(-2)/1)² – 2(4/1)] / (4/1)
= [(2)² – 8] / 4
= [4 – 8] / 4
= -4/4
= -1
Demikian
Semoga bermanfaat !