PSEB 11th Class Maths Solutions Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3

Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Ex 5.3

Question 1.
Solve the equation x² + 3 = 0
Answer.
The given quadratic equation is x² + 3 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 0, and c = 3
Therefore, the discriminant of the given equation is
D = b² – 4 ac
= 0² – 4 × 1 × 3 = – 12
Therefore, the required_solutions are

\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{\pm \sqrt{-12}}{2 \times 1}=\frac{\pm \sqrt{12} i}{2}\) [∵ √- 1 = i]

= \(\frac{\pm 2 \sqrt{3} i}{2}\) = ± √3 i.

Question 2.
Solve the equation 2x² + x + 1 = 0.
Answer.
The given quadratic equation is 2x² + x + 1=0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 2, b = 1 and c = 1
Therefore, the discriminant of the given equation is
D = b² – 4ac
= 1² – 4 × 2 × 1
= 1 – 8 = – 7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times 2}\)

= \(\frac{-1 \pm \sqrt{7} i}{4}\) [∵ √- 1 = i]

Question 3.
Solve the equation x² + r + 9 = 0
Ans.
The given equation x² + 3x + 9 =0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 3, and c = 9
Therefore, the discriminant of the given equation is
D = b² – 4 ac
= 3² – 4 × 1 × 9
= 9 – 36 = – 27
Therefore, the required_solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{-27}}{2(1)}\)

= \(\frac{-3 \pm 3 \sqrt{-3}}{2}=\frac{-3 \pm 3 \sqrt{3} i}{2}\) [∵ √- 1 = i].

Question 4.
Solve the equation – x² + x – 2 = 0.
Ans:
The given quadratic equation is – x² + x – 2 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = – 1, b = 1, and c = – 2
Therefore, the discriminant of the given equation is
D = b² – 4ac
= 1² – 4x(- 1) × (- 2)
= 1 – 8 = – 7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times(-1)}=\frac{-1 \pm \sqrt{7 i}}{-2}\) [∵ √- 1 = i]

Question 5.
Solve the equation x² + 3x + 5 = 0
Answer.
The given quadratic equation is x² + 3x + 5 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = 1, b = 3, and c = 5
Therefore, the discriminant of the given equation is
D = b² – 4ac
= 3² – 4 × 1 × 5
= 9 – 20 = – 11
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-3 \pm \sqrt{-11}}{2 \times 1}=\frac{-3 \pm \sqrt{11 i}}{2}\) [∵ √- 1 = i].

Question 6.
Solve the equation x² – x + 2 = 0
Answer.
The given quadratic equation is x² – x + 2 = 0
On comparing the given equation with ax²+ bx + c = 0, we obtain
a = 1 , b = – 1, and c = 2
Therefore, the discriminant of the given equation is
D = b² – 4ac
= (- 1)² – 4 × 1 × 2
= 1 – 8 = – 7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-1) \pm \sqrt{-7}}{2 \times 1}=\frac{1 \pm \sqrt{7} i}{2}\) [∵ √- 1 = i]

Question 7.
Solve the equation √2 x² + x + √2 = 0
Answer.
The given quadratic equation is √2 x² + x + √2 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = √2, b = 1, and c = √2
Therefore, the discriminant of the given equation is
D = b² – 4ac
= 1² – 4 × √2 × √2
= 1 – 8 = – 7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times \sqrt{2}}=\frac{-1 \pm \sqrt{7 i}}{2 \sqrt{2}}\) [∵ √- 1 = i]

Question 8.
Solve the equation √3x² – √2x + 3√3 = 0
Ans.
The given quadratic equation is √3x² – √2x + 3√3 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = √3, b = – √2, and c = 3√3
Therefore, the discriminant of the given equation is
D = b² – 4ac
= (- √2)² – 4(√3)(3√3)
= 2 – 36 = 34
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-(-\sqrt{2}) \pm \sqrt{-34}}{2 \times \sqrt{3}}\)

= \(\frac{\sqrt{2} \pm \sqrt{34} i}{2 \sqrt{3}}\) [∵ √- 1 = i]

Question 9.
Solve the equation x² + x + \(\frac{1}{\sqrt{\mathbf{2}}}=\) = 0
Ans.
We have, x² + x + \(\frac{1}{\sqrt{\mathbf{2}}}=\) = 0
√2x² + √2x + 1 = 0 …………….(i) [multiplying both sides by √2]
On Comparing eq. (i) with ax² + bx + c = 0, we get

Question 10.
Solve the equation x² + \(\frac{x}{\sqrt{2}}\) + 1 = 0
Ans.
The given quadratic equation is x² + \(\frac{x}{\sqrt{2}}\) + 1 = 0
The equation can also be written as √2x² + x + √2 = 0
On comparing the given equation with ax² + bx + c = 0, we obtain
a = √2, b = 1, and c = √2
∴ Discriminant(D) = b² – 4ac
= 1² – 4 × √2 × √2
= 1 – 8 = – 7
Therefore, the required solutions are
\(\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \sqrt{2}}=\frac{-1 \pm \sqrt{7 i}}{2 \sqrt{2}}\) [∵ √- 1 = i]