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 !