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In this lesson, we will learn how to solve quadratic equations with complex coefficients using the quadratic formula.

Q1:

Given that ( π₯ + 6 π π¦ ) + 4 0 ( 3 β π ) 3 + π = 0 2 , where π₯ and π¦ are real numbers, find all the possible values of π₯ and π¦ .

Q2:

Determine all the real values of π₯ and π¦ that satisfy the equation 8 6 π = ( π₯ β 2 2 π ) ( π¦ β π ) β 3 8 .

Q3:

Given that π₯ = β 4 + π is one of the roots of the equation 6 π₯ + 4 8 π₯ + π = 0 2 , find the other root and the value of π .

Q4:

Find the solution set of the equation in .

Q5:

Find the solution set of the equation ( 1 β π ) π₯ β ( 8 β 4 π ) π₯ + 5 + 7 π = 0 2 in β .

Q6:

Find the solution set of the equation ( 1 + π ) π₯ β ( 6 + 2 π ) π₯ + 3 β 5 π = 0 2 in β .

Q7:

Find the solution set of the equation ( 1 β π ) π₯ β ( 6 β 2 π ) π₯ + 7 + π = 0 2 in β .

Q8:

Find the solution set of the equation ( 1 β π ) π₯ β ( 6 β 2 π ) π₯ + 6 + 2 π = 0 2 in β .

Q9:

Find the solution set of the equation ( 1 β π ) π₯ β ( 6 β 2 π ) π₯ + 3 + 5 π = 0 2 in β .

Q10:

Find the solution set of the equation ( 1 + π ) π₯ β ( 6 + 2 π ) π₯ + 7 β π = 0 2 in β .

Q11:

Find the solution set of the equation ( 1 + π ) π₯ β ( 6 + 2 π ) π₯ + 6 β 2 π = 0 2 in β .

Q12:

Find the solution set of the equation ( 1 + π ) π₯ β ( 8 + 4 π ) π₯ + 5 β 7 π = 0 2 in β .

Q13:

Find all possible values of π§ , where π§ β β , for which 8 π§ = π§ + 1 2 2 .

Q14:

Find the solution set of ( π₯ + 6 ) β 2 ( π₯ + 6 ) + 1 = 0 ο¬ ο© in terms of π , where π is a complex cube root of unity.

Q15:

Given that ( π₯ + π¦ π ) = 2 β 2 π β 1 β π 2 , find all possible real values of π₯ and π¦ .

Q16:

Find the solution set of π₯ + 4 π π₯ β 4 π = 0 ο¨ in β , where π is a complex cube root of unity.

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