


Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5 (i) Deleted for CBSE Board 2022 Exams
Misc 5 (ii) Important
Misc 6
Misc 7
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11
Misc 12
Misc 13 Important Deleted for CBSE Board 2022 Exams
Misc 14 You are here
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Misc 19
Misc 20 Important
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