What is the answer to roots (x-3i)(x+3i) ?

The answer to roots (x-3i)(x+3i) is x=3i,x=-3i

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roots (x-3i)(x+3i)

roots

(x - 3 i) (x + 3 i)

x = 3 i, x = - 3 i

Solution steps

(x - 3 i) (x + 3 i)

The roots are the intercepts with the x-axis

(y = 0)

(x - 3 i) (x + 3 i) = 0

Substitute

x = a + bi

(a + bi - 3 i) (a + bi + 3 i) = 0

Expand

(a + bi - 3 i) (a + bi + 3 i) : (a^{2} - b^{2} + 9) + 2 iab

(a^{2} - b^{2} + 9) + 2 iab = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 9) + 2 iab = 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} + 9 = 0 2 ab = 0]

[a^{2} - b^{2} + 9 = 0 2 ab = 0] : (a = 0, a = 0, b = 3 b = - 3)

Substitute back

x = a + bi

x = 3 i, x = - 3 i

- n (δ + x y) x

π (4 - h^{2}) h

- \frac{t ( t - 50 ) ( t ^{2} - 150 t + 7500 )}{62500}

144 (\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = \frac{5}{4})

What is the answer to roots (x-3i)(x+3i) ?
The answer to roots (x-3i)(x+3i) is x=3i,x=-3i