Question 3, Exercise 1.3
Solutions of Question 3 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Question 3(i)
Solve the quadratic equation: 13z2+2z−16=0.
Solution. Given 13z2+2z−16=0⟹z2+6z−48=0 Apply the quadratic formula: z=−b±√b2−4ac2a, where a=1,b=6,andc=−48. Then z=−6±√36−4(1)(−48)2⋅1=−6±√2282⋅1=−6±2√572=−3±√57 Hence Solution set ={−3±√57}.
Question 3(ii)
Solve the quadratic equation: z2−12z+17=0.
Solution.
Given
z2−12z+17=0
Using the quadratic formula:
z=−b±√b2−4ac2a
Where
a=1,b=−12,c=17
Then
z=−(−12)±√(−12)2−4⋅1⋅172⋅1=12±√14−682=12±√1−27242=12±√−27142=12±√−27122=1±√271i4
Therefore, the solution set is: {1±√271i4}
Question 3(iii)
Solve the quadratic equation: z2−6z+25=0.
Solution.
Given z2−6z+25=0 Using the quadratic formula: z=−b±√b2−4ac2a Where a=1,b=−6,c=25 Then z=−(−6)±√(−6)2−4⋅1⋅252⋅1=6±√36−1002=6±√−642=6±8i2=3±4i
Therefore, the solution set is: {3±4i}
Question 3(iv)
Solve the quadratic equation: z2−9z+11=0.
Solution.
Given z2−9z+11=0 Using the quadratic formula: z=−b±√b2−4ac2a
Where a=1,b=−9,c=11 Then z=−(−9)±√(−9)2−4⋅1⋅112⋅1=9±√81−442=9±√372
Therefore, the solution set ={9±√372}
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