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Last updated at May 29, 2018 by Teachoo

Misc 14 Find the real numbers x and y if (π₯ β ππ¦) (3 + 5π) is the conjugate of β6 β 24π. Conjugate of β6 β24π = β 6 + 24π Now it is given that (π₯ β ππ¦) (3 + 5π) is conjugate of β6 + 24π Hence from (1) and (2) β 6 + 24π = (π₯ β ππ¦) (3 + 5π) β 6 + 24π = π₯ ( 3 + 5π ) β ππ¦ ( 3 + 5π) β 6 + 24π = 3π₯ + 5π₯π β 3π¦π β 5π2π¦ Putting π2 = β1 β 6 + 24π = 3π₯ + 5π₯π β 3π¦π β 5 Γ β 1 Γ π¦ β 6 + 24π = 3π₯ + 5π₯π β 3π¦π + 5π¦ β 6 + 24π = 3π₯ + 5π¦ β 3π¦π + 5π₯π β 6 + 24π = 3π₯ + 5π¦ + ( β 3π¦ + 5π₯)π Comparing real parts β 6 = 3π₯ + 5π¦ Comparing imaginary parts 24 = 5π₯ β 3π¦ We solve equation (3) and (4) to find the value of x and y From (3) 3π₯ + 5π¦ = β6 3π₯ = β6 β 5π¦ π₯ = (β6 β 5π¦)/3 Putting π₯ = (β6 β 5π¦)/3 in (4) 5π₯ β 3π¦ = 24 5 ((β6 β 5π¦)/3) β 3π¦ = 24 Multiplying 3 both sides 3Γ5((β6 β 5π¦)/3) β 3Γ3π¦ = 3Γ24 5(β6 β 5π¦) β 9π¦ = 72 5 (β 6) β 5 (5π¦) β 9π¦ = 72 β30 β 25π¦β 9π¦ = 72 β 25π¦ β 9π¦ = 72 + 30 β 34π¦ = 102 π¦ = 102/(β34) π¦ = β3 Putting y = β 3 in (4) 24 = 5π₯ β 3π¦ 24 = 5π₯ β 3(β3) 24 = 5π₯+9 24 β9= 5π₯ 15= 5π₯ 5π₯=15 π₯ = 15/5 π₯ = 3 Hence value of π₯ = 3 and π¦ = β3