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受欢迎的代数 >

2^2+sqrt(-1)+x^2=2

解答

2^{2} + - 1 + x^{2} = 2

解答

x = 45 cos (\frac{arctan ( \frac{1}{2} ) - π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) - π}{2}), x = 45 cos (\frac{arctan ( \frac{1}{2} ) + π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) + π}{2})

求解步骤

2^{2} + - 1 + x^{2} = 2

将

- 1

到右边

2^{2} + x^{2} = 2 - i

将

2^{2}

到右边

x^{2} = - 2 - i

For

z^{n} = a

the solutions are

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

对于

n = 2, a = - 2 - i

∣ a ∣ = 5

ar g (a) = arctan (\frac{1}{2}) - π

x = 5 (cos (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 0 π}{2}) + i sin (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 0 π}{2})), x = 5 (cos (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 1 π}{2}) + i sin (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 1 π}{2}))

化简

5 (cos (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 0 π}{2}) + i sin (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 0 π}{2})) : 45 cos (\frac{arctan ( \frac{1}{2} ) - π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) - π}{2})

化简

5 (cos (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 1 π}{2}) + i sin (\frac{arctan ( \frac{1}{2} ) - π + 2 \cdot 1 π}{2})) : 45 cos (\frac{arctan ( \frac{1}{2} ) + π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) + π}{2})

x = 45 cos (\frac{arctan ( \frac{1}{2} ) - π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) - π}{2}), x = 45 cos (\frac{arctan ( \frac{1}{2} ) + π}{2}) + 45 i sin (\frac{arctan ( \frac{1}{2} ) + π}{2})

作图

流行的例子

125 x^{2} + 20 = 0

solvefor x,9x^2+16y^2=0

so l v e f or x, 9 x^{2} + 16 y^{2} = 0

l(q)=5q^2-8q+12

l (q) = 5 q^{2} - 8 q + 12

solvefor t,dt+0.5t=dt(0.5t)

so l v e f or t, d t + 0.5 t = d t (0.5 t)

(7-h)^2+(3-k)^2=r^2

(7 - h)^{2} + (3 - k)^{2} = r^{2}