Question

Solve: 2x2 -(3 + 7i) x - (3 - 9i) = 0​

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: quadratic \: equation \: is \\ 2x {}^{2} - (3 + 7i)x -( 3 - 9i) = 0 \\ \\ comparing \: it \: with \: ax {}^{2} + bx + c = 0 \\ we \: get \\ a = 2 \: ,b = - (3 + 7i) \: ,c = - (3 - 9i) \\ \\ so \: applying \\ Δ = b {}^{2} -4 ac \\ = ( - (3 + 7i)) {}^{2} \: - \: 4 \times 2 \times( -( 3 - 9i)) \\ = (3 + 7i) {}^{2} + 8(3 - 9i) \\ = 9 + 49i {}^{2} + 42i + 24 - 72i \\ = 33 + 49i {}^{2} - 30i \\ = 33 + 49( - 1) - 30i \\ = 33 - 49 - 30i \\ - 16 - 30i \\ = - 30i - 16 \\ \\ now \: by \: using \: formula \: method \\ we \: have \\ \\ x = \frac{ - b \:± \: \sqrt{b {}^{2} - 4ac } }{2a } \\ \\ = \frac{ - - (3 + 7i) \: ± \: \sqrt{ - 30i - 16} }{(2 \times 2)} \\ \\ = \frac{3 + 7i \:± \: \sqrt{ - 30i - 16} }{4}$

$so \: let \: then \\ \\ \sqrt{ - 16 - 30i} = a + ib \\ \\ squaring \: both \: sides \\ we \: get \\ \\ - 16 - 30i = a {}^{2} + bi {}^{2} + 2aib \\ \\ - 16 - 30i = a {}^{2} + b {}^{2} ( - 1) + 2aib \\ \\ = a {}^{2} - b {}^{2} + 2aib \\ \\ comparing \: real \: and \: imaginary \: parts \\ we \: get \\ \\ a {}^{2} - b {}^{2} = - 16 \: \: \: \: \: \: (1) \\ \\ 2ab = - 30 \\ \\ ab = - 15 \\ \\ a = \frac{ - 15}{b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

$substitute \: value \: of \: a \: in \: (1) \\ \\ a {}^{2} - b {}^{2} = - 16 \\ \\ ( \frac{ - 15}{b} ) {}^{2} - b {}^{2} = - 16 \\ \\ ( \frac{225}{b {}^{2} }) - b {}^{2} = - 16 \\ \\ let \: then \: \\ b {}^{2} = x \\ \\ hence \: accordingly \\ \\ \frac{225}{x} - x = - 16 \\ \\ 225 - x {}^{2} = - 16x \\ \\ x {}^{2} - 16x - 225 = 0 \\ \\ x {}^{2} - 25x + 9x - 225 = 0 \\ \\( x + 9)(x - 25) = 0 \\ \\ x = 25 \: \: or \: \: x = - 9 \\ \\ but \: if \: we \: take \: x = - 9 \\ \\ b {}^{2} = - 9 \\ \\ b {}^{2} = \sqrt{ - 9} \: \: ∈ \: \: i \\ \\ hence \: \: as \: a,b \: ∈ \: \: R \\ \\ x = - 9 \: \: is \: absurd \\ thus \: x = 25$

$hence \: then \\ b {}^{2} = 25 \\ b = ±5 \\ \\ therefore \: then \\ \\ a = ±3 \\ hence \: considering \: \\ (a ,b) = (3 , - 5) \\ \\ so \: as \: \\ \\ x = \frac{3 + 7i \:± \: ( - 16 - 30i) }{4} \\ \\ = \frac{3 + 7i \: ± \: (3 - 5i)}{4} \\ \\ x = \frac{3 + 7i + (3 - 5i)}{4} \: \: or \: \: x = \frac{3 + 7i \: - (3 - 5i)}{4} \\ \\ = \frac{6 + 2i}{4} \: \: or \: \: = \frac{7i + 5i}{4} \\ \\ = \frac{2(3 + i)}{4} \: \: \: or \: \: = \frac{12i}{4} \\ \\ x = \frac{3 + i}{2} \: \: \: or \: \: \: x = 3i$