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人気のある代数 >

(2x^2-4)^2+9=0

解

(2 x^{2} - 4)^{2} + 9 = 0

解

x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} ) + 2 π}{2}), x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) - i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2})

解答ステップ

(2 x^{2} - 4)^{2} + 9 = 0

9

を右側に移動します

(2 x^{2} - 4)^{2} = - 9

(g (x))^{2} = f (a)

の場合, 解は

g (x) = f (a), - f (a)

解く

2 x^{2} - 4 = - 9 : x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} ) + 2 π}{2})

解く

2 x^{2} - 4 = - - 9 : x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) - i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2})

解答は

x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} ) + 2 π}{2}), x = \frac{5}{2} cos (\frac{arctan ( \frac{3}{4} )}{2}) - i \frac{5}{2} sin (\frac{arctan ( \frac{3}{4} )}{2}), x = \frac{5}{2} cos (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2}) + i \frac{5}{2} sin (\frac{- arctan ( \frac{3}{4} ) + 2 π}{2})

グラフ

人気の例

1 - (4 - 3 x)^{3} = 9

16x^4-40x^2+9=0

16 x^{4} - 40 x^{2} + 9 = 0

2 x^{4} + 4 x - 6 = 0

(x+2)(x+3)(x+4)(x+5)=195

(x + 2) (x + 3) (x + 4) (x + 5) = 195

x^{10} = 0.9