What is the answer to x^2-(7+3i)x+(10+11i)=0 ?

The answer to x^2-(7+3i)x+(10+11i)=0 is x=3+2i,x=4+i

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x^2-(7+3i)x+(10+11i)=0

x^{2} - (7 + 3 i) x + (10 + 11 i) = 0

x = 3 + 2 i, x = 4 + i

Solution steps

x^{2} - (7 + 3 i) x + (10 + 11 i) = 0

Substitute

x = a + bi

(a + bi)^{2} - (7 + 3 i) (a + bi) + 10 + 11 i = 0

Expand

(a + bi)^{2} - (7 + 3 i) (a + bi) + 10 + 11 i : (a^{2} - b^{2} - 7 a + 3 b + 10) + i (11 - 3 a - 7 b + 2 ab)

(a^{2} - b^{2} - 7 a + 3 b + 10) + i (11 - 3 a - 7 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - 7 a + 3 b + 10) + i (11 - 3 a - 7 b + 2 ab) = 0 + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} - 7 a + 3 b + 10 = 0 11 - 3 a - 7 b + 2 ab = 0]

[a^{2} - b^{2} - 7 a + 3 b + 10 = 0 11 - 3 a - 7 b + 2 ab = 0] : (a = 3, a = 4, b = 2 b = 1)

Substitute back

x = a + bi

x = 3 + 2 i, x = 4 + i

∣ a - 1 + 2 i ∣ = 3

z - (1 + i) w = 2

6 + 7 i = x + y i

\frac{4 i}{- 6 + 4 i} = \frac{a}{b}

\frac{15}{z - 1} = - 6 - 3 i

What is the answer to x^2-(7+3i)x+(10+11i)=0 ?
The answer to x^2-(7+3i)x+(10+11i)=0 is x=3+2i,x=4+i