ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある代数 >

sqrt(-1)=x^2+4

解

- 1 = x^{2} + 4

解

x = 417 cos (\frac{- arctan ( \frac{1}{4} ) + π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + π}{2}), x = 417 cos (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2})

解答ステップ

- 1 = x^{2} + 4

拡張

- 1 : i

i = x^{2} + 4

辺を交換する

x^{2} + 4 = i

4

を右側に移動します

x^{2} = i - 4

z^{n} = a

の場合, 解は

z_{k} = n ∣ a ∣ (cos (\frac{ar g ( a ) + 2 kπ}{n}) + i sin (\frac{ar g ( a ) + 2 kπ}{n})),

k = 0, 1, \dots, n - 1

以下のため:

n = 2, a = i - 4

∣ a ∣ = 17

ar g (a) = - arctan (\frac{1}{4}) + π

x = 17 (cos (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 0 π}{2}) + i sin (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 0 π}{2})), x = 17 (cos (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 1 π}{2}) + i sin (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 1 π}{2}))

簡素化

17 (cos (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 0 π}{2}) + i sin (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 0 π}{2})) : 417 cos (\frac{- arctan ( \frac{1}{4} ) + π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + π}{2})

簡素化

17 (cos (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 1 π}{2}) + i sin (\frac{- arctan ( \frac{1}{4} ) + π + 2 \cdot 1 π}{2})) : 417 cos (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2})

x = 417 cos (\frac{- arctan ( \frac{1}{4} ) + π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + π}{2}), x = 417 cos (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2}) + 417 i sin (\frac{- arctan ( \frac{1}{4} ) + 3 π}{2})

グラフ

人気の例

x^{2} = - 16

根 x^2+8x+25=0

roo t s x^{2} + 8 x + 25 = 0

8 x^{2} + 9 = 313

3 x^{2} + 18 = 15 x

x^{2} - 2 x - 3 = - 4