Ada yang bisa bantu?;).... Question from @hanyardaa

Takamori37 Selisih kuadrat akar-akar persamaan kuadratnya adalah 20.
x₁²-x₂² = 20
Maka, untuk menyederhanakan (mempermudah)
Kuadratkan:
(x₁²-x₂²)² = 400
x₁⁴-2x₁²x₂²+x₂⁴ = 400
(x₁⁴+x₂⁴)-2(x₁x₂)² = 400
Fragmentasikan:
(x₁²+x₂²)²-4(x₁x₂)² = 400
[(x₁+x₂)²-2x₁x₂]² - 4(x₁x₂)² = 400
Masukkan ke persamaan:
[(2m+4)²-16m]² - 4[8m]² = 400
[4m²+16]² - 256m² = 400
16m⁴ + 128m² + 256 - 256m² = 400
16m⁴ - 128m² - 144 = 0
m⁴ - 8m² - 9 = 0
(m² + 1)(m² - 9) = 0

Untuk solusi realnya, didapat m² = 9
Dihasilkan:
m² - 4
= 9 - 4
= 5 [D]

$\boxed{\boxed{Mr}}\boxed{Crussader}$

1 votes Thanks 1

izharulhaq78 $\displaystyle x_1^2-x_2^2=20\\(x_1-x_2)(x_1+x_2)=20\\(\frac{\sqrt{D}}{a})(\frac{-b}{a})=20\\\frac{\sqrt{D}(-b)}{a^2}=20\\\frac{(\sqrt{b^2-4ac})(-b)}{a^2}=20\\\frac{(\sqrt{(-2m-4)^2-4(1)(8m)})(-(-2m-4))}{1^2}=20\\(\sqrt{4m^2+16m+16-32m})(2m+4)=20\\(\sqrt{4m^2-16m+16})(2m+4)=20\\(\sqrt{4(m^2-4m+4)})2(m+2)=20\\2(\sqrt{m^2-4m+4})(m+2)=10\\(\sqrt{(m-2)^2})(m+2)=5\\(m-2)(m+2)=5\\\boxed{\boxed{m^2-4=5}}$

0 votes Thanks 1

Suatu oksida (Mr=160) mengandung 70% massa unsur X. Jika massa atom relatif O=16, rumus molekul oksida tersebut adalah... a.XO b.XO2 c.X3O d.X2O e.X2O3

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