Question

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Answer 1

$\large\underline\mathfrak\red{Given \: :- }$

• p and q are real.
• x² + (p-iq)x + 3i = 0 is 8.

$\large\underline\mathfrak\red{To \: find \: :- }$

• p and q

$\large\underline\mathfrak\red{Solution \: :- }$

We know that,

$\sf{x {}^{2} + (p + iq)x + 3i = 0 }$

$\sf{(α+β) = -(q-iq),αβ=3i}$

Squaring,

$\sf{α²+β² = (α+β)² -2αβ = [-(p+iq)] -6i }$

Required

$\sf\implies{(p {}^{2} - q {}^{2}) + i(2pq - 6) = 8}$

$\sf\implies{p {}^{2} - q {}^{2} = 8 \: and \: pq = 3 }$

$\sf\implies{p = 3,q = 1 \: \: or \: \: p = - 3,q = - 1}$

Values :-

p = 3 , q = 1 or p = -3 , q = -1