Maharashtra Board Class 11 Maths Part 1 Chapter 3 Complex Numbers 3.2 Solutions

Question 1. Find the square root of the following complex numbers:
(i) -8 - 6i
Answer:
Let \( \sqrt{-8 - 6i} = a + bi \), where \( a, b \in R \).
Squaring on both sides, we get:
\( -8 - 6i = (a + bi)^2 \)
\( -8 - 6i = a^2 + b^2 i^2 + 2abi \)
\( -8 - 6i = (a^2 - b^2) + 2abi \) .....[\( \because i^2 = -1 \)]
By equating the real and imaginary parts, we can set up a system of simultaneous equations to solve for the unknown real numbers a and b. This systematic algebraic approach ensures we find both possible square roots of the given complex number.
In simple words: To find the square root of a complex number, we assume the answer is another complex number \( a + bi \). We square both sides and match the real parts together and the imaginary parts together to solve for \( a \) and \( b \).

🎯 Exam Tip: Always remember that \( i^2 = -1 \) when expanding the squared term, which changes the sign of the \( b^2 \) term to negative.

Question 1. Find the square root of the following complex numbers:
(i) \( -8 - 6i \)
(ii) \( 7 + 24i \)
Answer:
(i) Let \( \sqrt{-8 - 6i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get:
\( -8 - 6i = (a + bi)^2 \)
\( -8 - 6i = a^2 + b^2i^2 + 2abi \)
\( -8 - 6i = (a^2 - b^2) + 2abi \) [since \( i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = -8 \) and \( 2ab = -6 \)
\( \implies a^2 - b^2 = -8 \) and \( b = \frac{-3}{a} \)
\( \implies a^2 - \left(\frac{-3}{a}\right)^2 = -8 \)
\( \implies a^2 - \frac{9}{a^2} = -8 \)
\( \implies a^4 - 9 = -8a^2 \)
\( \implies a^4 + 8a^2 - 9 = 0 \)
\( \implies (a^2 + 9)(a^2 - 1) = 0 \)
\( \implies a^2 = -9 \) or \( a^2 = 1 \)
But \( a \in \mathbb{R} \)
\( \implies a^2 \neq -9 \)
\( \implies a^2 = 1 \)
\( \implies a = \pm 1 \)
When \( a = 1 \), \( b = \frac{-3}{1} = -3 \)
When \( a = -1 \), \( b = \frac{-3}{-1} = 3 \)
\( \implies \sqrt{-8 - 6i} = \pm(1 - 3i) \)

(ii) Let \( \sqrt{7 + 24i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get:
\( 7 + 24i = (a + bi)^2 \)
\( 7 + 24i = a^2 + b^2i^2 + 2abi \)
\( 7 + 24i = (a^2 - b^2) + 2abi \) [since \( i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 7 \) and \( 2ab = 24 \)
\( \implies a^2 - b^2 = 7 \) and \( b = \frac{12}{a} \)
\( \implies a^2 - \left(\frac{12}{a}\right)^2 = 7 \)
\( \implies a^2 - \frac{144}{a^2} = 7 \)
\( \implies a^4 - 144 = 7a^2 \)
\( \implies a^4 - 7a^2 - 144 = 0 \)
\( \implies (a^2 - 16)(a^2 + 9) = 0 \)
\( \implies a^2 = 16 \) or \( a^2 = -9 \)
But \( a \in \mathbb{R} \)
\( \implies a^2 \neq -9 \)
\( \implies a^2 = 16 \)
\( \implies a = \pm 4 \)
When \( a = 4 \), \( b = \frac{12}{4} = 3 \)
When \( a = -4 \), \( b = \frac{12}{-4} = -3 \)
\( \implies \sqrt{7 + 24i} = \pm(4 + 3i) \)
In simple words: To find the square root of a complex number, we set it equal to \( a + bi \), square both sides, and solve for the real numbers \( a \) and \( b \) by comparing the real and imaginary parts.

🎯 Exam Tip: Always state clearly that \( a \in \mathbb{R} \) so you can discard the negative value of \( a^2 \). This step is crucial for securing full marks from examiners.

Question 1. Find the square root of the following complex number:
(ii) \( 7 + 24i \)
Answer:
Let \( \sqrt{7 + 24i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring both sides, we get:
\( 7 + 24i = (a + bi)^2 \)
\( 7 + 24i = a^2 + b^2i^2 + 2abi \)
\( 7 + 24i = (a^2 - b^2) + 2abi \) [Since \( i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 7 \) and \( 2ab = 24 \)

\( \implies b = \frac{12}{a} \)
Substituting the value of \( b \) in \( a^2 - b^2 = 7 \):
\( a^2 - \left(\frac{12}{a}\right)^2 = 7 \)

\( \implies a^2 - \frac{144}{a^2} = 7 \)

\( \implies a^4 - 144 = 7a^2 \)

\( \implies a^4 - 7a^2 - 144 = 0 \)

\( \implies (a^2 - 16)(a^2 + 9) = 0 \)

\( \implies a^2 = 16 \) or \( a^2 = -9 \)
But \( a \in \mathbb{R} \), therefore \( a^2 \neq -9 \)

\( \implies a^2 = 16 \)

\( \implies a = \pm 4 \)
When \( a = 4 \), \( b = \frac{12}{4} = 3 \)
When \( a = -4 \), \( b = \frac{12}{-4} = -3 \)

\( \implies \sqrt{7 + 24i} = \pm(4 + 3i) \)
In simple words: To find the square root, we set up equations by comparing the real and imaginary parts, solve for the real numbers \( a \) and \( b \), and write the final answer in the form \( \pm(a + bi) \).

🎯 Exam Tip: Always write the condition \( a \in \mathbb{R} \) clearly to justify why you are discarding the negative value of \( a^2 \).

Question 1. Find the square root of the following complex number:
(iii) \( 1 + 4\sqrt{3}i \)
Answer:
Let \( \sqrt{1 + 4\sqrt{3}i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring both sides, we get:
\( 1 + 4\sqrt{3}i = (a + bi)^2 \)
\( 1 + 4\sqrt{3}i = a^2 + b^2i^2 + 2abi \)
\( 1 + 4\sqrt{3}i = (a^2 - b^2) + 2abi \) [Since \( i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 1 \) and \( 2ab = 4\sqrt{3} \)

\( \implies b = \frac{2\sqrt{3}}{a} \)
Substituting the value of \( b \) in \( a^2 - b^2 = 1 \):
\( a^2 - \left(\frac{2\sqrt{3}}{a}\right)^2 = 1 \)

\( \implies a^2 - \frac{12}{a^2} = 1 \)

\( \implies a^4 - 12 = a^2 \)

\( \implies a^4 - a^2 - 12 = 0 \)

\( \implies (a^2 - 4)(a^2 + 3) = 0 \)

\( \implies a^2 = 4 \) or \( a^2 = -3 \)
But \( a \in \mathbb{R} \), therefore \( a^2 \neq -3 \)

\( \implies a^2 = 4 \)

\( \implies a = \pm 2 \)
When \( a = 2 \), \( b = \frac{2\sqrt{3}}{2} = \sqrt{3} \)
When \( a = -2 \), \( b = \frac{2\sqrt{3}}{-2} = -\sqrt{3} \)

\( \implies \sqrt{1 + 4\sqrt{3}i} = \pm(2 + \sqrt{3}i) \)
In simple words: We assume the square root is \( a + bi \), square both sides to compare real and imaginary parts, and solve the quadratic equation to find the values of \( a \) and \( b \).

🎯 Exam Tip: Be careful when squaring terms with square roots, such as \( (2\sqrt{3})^2 = 4 \times 3 = 12 \), to avoid simple calculation errors.

Question (iii) \( 1 + 4\sqrt{3}i \)
Answer: Equating real and imaginary parts, we get:
\( a^2 - b^2 = 1 \) and \( 2ab = 4\sqrt{3} \)
\( \implies a^2 - b^2 = 1 \) and \( b = \frac{2\sqrt{3}}{a} \)
\( \implies a^2 - \left(\frac{2\sqrt{3}}{a}\right)^2 = 1 \)
\( \implies a^2 - \frac{12}{a^2} = 1 \)
\( \implies a^4 - 12 = a^2 \)
\( \implies a^4 - a^2 - 12 = 0 \)
\( \implies (a^2 - 4)(a^2 + 3) = 0 \)
\( \implies a^2 = 4 \) or \( a^2 = -3 \)
But since \( a \) is a real number, we have \( a \in \mathbb{R} \). This step-by-step algebraic simplification helps us find the exact real values of our variables.
\( \implies a^2 \neq -3 \)
\( \implies a^2 = 4 \)
\( \implies a = \pm 2 \)
When \( a = 2 \), \( b = \frac{2\sqrt{3}}{2} = \sqrt{3} \)
When \( a = -2 \), \( b = \frac{2\sqrt{3}}{-2} = -\sqrt{3} \)
\( \implies \sqrt{1+4\sqrt{3}i} = \pm (2 + \sqrt{3}i) \)
In simple words: To find the square root of a complex number, we set it equal to \( a + bi \), square both sides, and solve for the real numbers \( a \) and \( b \).

🎯 Exam Tip: Remember that since \( a \) must be a real number, \( a^2 \) cannot be negative. Always discard the negative value of \( a^2 \) with a proper reason to avoid losing marks.

Question (iv) \( 3 + 2\sqrt{10}i \)
Answer: Let \( \sqrt{3 + 2\sqrt{10}i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get:
\( 3 + 2\sqrt{10}i = (a + bi)^2 \)
\( \implies 3 + 2\sqrt{10}i = a^2 + b^2i^2 + 2abi \)
\( \implies 3 + 2\sqrt{10}i = (a^2 - b^2) + 2abi \) ..... [\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get:
\( a^2 - b^2 = 3 \) and \( 2ab = 2\sqrt{10} \)
\( \implies ab = \sqrt{10} \)
\( \implies b = \frac{\sqrt{10}}{a} \)
Substitute \( b = \frac{\sqrt{10}}{a} \) in \( a^2 - b^2 = 3 \):
\( \implies a^2 - \left(\frac{\sqrt{10}}{a}\right)^2 = 3 \)
\( \implies a^2 - \frac{10}{a^2} = 3 \)
\( \implies a^4 - 10 = 3a^2 \)
\( \implies a^4 - 3a^2 - 10 = 0 \)
\( \implies (a^2 - 5)(a^2 + 2) = 0 \)
Since \( a \in \mathbb{R} \), \( a^2 \geq 0 \), so \( a^2 \neq -2 \). This systematic method ensures we do not miss any of the possible roots.
\( \implies a^2 = 5 \)
\( \implies a = \pm \sqrt{5} \)
When \( a = \sqrt{5} \), \( b = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2} \)
When \( a = -\sqrt{5} \), \( b = \frac{\sqrt{10}}{-\sqrt{5}} = -\sqrt{2} \)
\( \implies \sqrt{3 + 2\sqrt{10}i} = \pm (\sqrt{5} + \sqrt{2}i) \)
In simple words: We find the square root by setting up equations for the real and imaginary parts, solving for \( a \) and \( b \), and writing the final answer in the form \( \pm(a + bi) \).

🎯 Exam Tip: Always write the final answer with a \( \pm \) sign because every non-zero complex number has two square roots.

Question 1. Find the square root of \( 3 + 2\sqrt{10}i \).
Answer:
We have \( a^2 - b^2 = 3 \) and \( b = \frac{\sqrt{10}}{a} \).
\( \therefore a^2 - \left(\frac{\sqrt{10}}{a}\right)^2 = 3 \)
\( \dots a^2 - \frac{10}{a^2} = 3 \)
\( \therefore a^4 - 10 = 3a^2 \)
\( \therefore a^4 - 3a^2 - 10 = 0 \)
\( \therefore (a^2 - 5)(a^2 + 2) = 0 \)
\( \therefore a^2 = 5 \) or \( a^2 = -2 \)
But \( a \in \mathbb{R} \)
\( \therefore a^2 \neq -2 \)
\( \therefore a^2 = 5 \)
\( \therefore a = \pm\sqrt{5} \)
When \( a = \sqrt{5} \), \( b = \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2} \)
When \( a = -\sqrt{5} \), \( b = \frac{\sqrt{10}}{-\sqrt{5}} = -\sqrt{2} \)
\( \therefore \sqrt{3 + 2\sqrt{10}i} = \pm(\sqrt{5} + \sqrt{2}i) \)
In simple words: We solve the equations for \( a \) and \( b \) by substituting \( b \) into the first equation, factoring the quadratic expression, and keeping only the real values of \( a \) to find the final square root.

🎯 Exam Tip: Remember that since \( a \) is a real number, \( a^2 \) cannot be negative. Always discard the negative value of \( a^2 \) to avoid losing marks on invalid imaginary roots.

Question 2. Find the square root of the following complex number:
(v) \( 2(1 - \sqrt{3}i) \)
Answer:
Let \( \sqrt{2(1 - \sqrt{3}i)} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 2(1 - \sqrt{3}i) = (a + bi)^2 \)
\( 2(1 - \sqrt{3}i) = a^2 + b^2i^2 + 2abi \)
\( 2 - 2\sqrt{3}i = (a^2 - b^2) + 2abi \quad \dots [\because i^2 = -1] \)
In simple words: To find the square root of a complex number, we assume it equals \( a + bi \), square both sides, and then equate the real and imaginary parts to solve for \( a \) and \( b \).

🎯 Exam Tip: When squaring \( a + bi \), always remember that \( i^2 = -1 \), which changes the sign of the \( b^2 \) term to \( -b^2 \).

Question 1. Find the square root of \( 2(1 - \sqrt{3}i) \).
Answer: Let the square root of \( 2(1 - \sqrt{3}i) \) be \( a + bi \), where \( a, b \in \mathbb{R} \). Squaring both sides and equating real and imaginary parts, we get:
\( a^2 - b^2 = 2 \) and \( 2ab = -2\sqrt{3} \)
\( \implies a^2 - b^2 = 2 \) and \( b = -\frac{\sqrt{3}}{a} \)
\( \implies a^2 - \left(-\frac{\sqrt{3}}{a}\right)^2 = 2 \)
\( \implies a^2 - \frac{3}{a^2} = 2 \)
\( \implies a^4 - 3 = 2a^2 \)
\( \implies a^4 - 2a^2 - 3 = 0 \)
\( \implies (a^2 - 3)(a^2 + 1) = 0 \)
\( \implies a^2 = 3 \) or \( a^2 = -1 \)
But since \( a \) is a real number, \( a \in \mathbb{R} \).
\( \implies a^2 \neq -1 \)
\( \implies a^2 = 3 \)
\( \implies a = \pm\sqrt{3} \)
When \( a = \sqrt{3} \), \( b = \frac{-\sqrt{3}}{\sqrt{3}} = -1 \)
When \( a = -\sqrt{3} \), \( b = \frac{-\sqrt{3}}{-\sqrt{3}} = 1 \)
\( \implies \sqrt{2(1-\sqrt{3}i)} = \pm(\sqrt{3} - i) \)
This elegant method allows us to find the roots of any complex number by converting the problem into a system of real equations.
In simple words: To find the square root of a complex number, we set it equal to \( a + bi \), square both sides, and solve for the real numbers \( a \) and \( b \).

🎯 Exam Tip: Always remember that since \( a \) is a real number, its square cannot be negative, which allows you to discard the negative value of \( a^2 \) immediately.

Question 2. Solve the following quadratic equations.
(i) \( 8x^2 + 2x + 1 = 0 \)
Answer:
(i) Given equation is \( 8x^2 + 2x + 1 = 0 \)
Comparing with the standard form \( ax^2 + bx + c = 0 \), we get:
\( a = 8, b = 2, c = 1 \)
Discriminant \( = b^2 - 4ac \)
\( = (2)^2 - 4 \times 8 \times 1 \)
\( = 4 - 32 \)
\( = -28 < 0 \)
Since the discriminant is negative, the given equation has complex roots.
These roots are given by the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \implies x = \frac{-2 \pm \sqrt{-28}}{2(8)} \)
\( \implies x = \frac{-2 \pm 2\sqrt{7}i}{16} \)
\( \implies x = \frac{-1 \pm \sqrt{7}i}{8} \)
Thus, the complex roots of the quadratic equation are \( \frac{-1 + \sqrt{7}i}{8} \) and \( \frac{-1 - \sqrt{7}i}{8} \). This shows how quadratic equations can have solutions even when their graphs do not cross the x-axis.
In simple words: When the discriminant (the value under the square root) is negative, the equation has no real solutions, only complex solutions containing the imaginary unit \( i \).

🎯 Exam Tip: Always write down the values of \( a \), \( b \), and \( c \) clearly before substituting them into the discriminant formula to avoid simple calculation errors.

Question 1. Solve the following quadratic equations:
(i) \( 8x^2 + 2x + 1 = 0 \)
(ii) \( 2x^2 - \sqrt{3}x + 1 = 0 \)
(iii) \( 3x^2 - 7x + 5 = 0 \)
Answer:
(i) \( 8x^2 + 2x + 1 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 8, b = 2, c = 1 \)
Discriminant \( = b^2 - 4ac \)
\( = (2)^2 - 4 \times 8 \times 1 \)
\( = 4 - 32 \)
\( = -28 < 0 \)
So, the given equation has complex roots.
These roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-2 \pm \sqrt{-28}}{2(8)} \)
\( = \frac{-2 \pm 2\sqrt{7}i}{16} \)
\( x = \frac{-1 \pm \sqrt{7}i}{8} \)
\( \therefore \) the roots of the given equation are \( \frac{-1+\sqrt{7}i}{8} \) and \( \frac{-1-\sqrt{7}i}{8} \)

(ii) \( 2x^2 - \sqrt{3}x + 1 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 2, b = -\sqrt{3}, c = 1 \)
Discriminant \( = b^2 - 4ac \)
\( = (-\sqrt{3})^2 - 4 \times 2 \times 1 \)
\( = 3 - 8 \)
\( = -5 < 0 \)
So, the given equation has complex roots.
These roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-\sqrt{3}) \pm \sqrt{-5}}{2(2)} \)
\( x = \frac{\sqrt{3} \pm \sqrt{5}i}{4} \)
\( \therefore \) the roots of the given equation are \( \frac{\sqrt{3}+\sqrt{5}i}{4} \) and \( \frac{\sqrt{3}-\sqrt{5}i}{4} \)

(iii) \( 3x^2 - 7x + 5 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 3, b = -7, c = 5 \)
Discriminant \( = b^2 - 4ac \)
\( = (-7)^2 - 4 \times 3 \times 5 \)
\( = 49 - 60 \)
\( = -11 < 0 \)
So, the given equation has complex roots.
These roots are given by
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-(-7) \pm \sqrt{-11}}{2(3)} \)
\( x = \frac{7 \pm \sqrt{11}i}{6} \)
\( \therefore \) the roots of the given equation are \( \frac{7+\sqrt{11}i}{6} \) and \( \frac{7-\sqrt{11}i}{6} \)
In simple words: To solve these quadratic equations, we first find the discriminant. Since the discriminant is negative, the equations have complex roots containing the imaginary unit \( i \), which we find using the quadratic formula.

🎯 Exam Tip: When the discriminant \( b^2 - 4ac \) is negative, remember to replace \( \sqrt{-d} \) with \( \sqrt{d}i \) to express the roots in complex form \( a + bi \). Always simplify the final fraction to its lowest terms to secure full marks.

Question 2. Solve the quadratic equation:
(iii) \( 3x^2 - 7x + 5 = 0 \)
Answer:
Given equation is \( 3x^2 - 7x + 5 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get:
\( a = 3, b = -7, c = 5 \)
Discriminant \( = b^2 - 4ac \)
\( = (-7)^2 - 4 \times 3 \times 5 \)
\( = 49 - 60 \)
\( = -11 < 0 \)
So, the given equation has complex roots. These roots are conjugate pairs of each other.
These roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ = \frac{-(-7) \pm \sqrt{-11}}{2(3)} \]
\[ = \frac{7 \pm \sqrt{11}i}{6} \]
\( \therefore \) the roots of the given equation are \( \frac{7+\sqrt{11}i}{6} \) and \( \frac{7-\sqrt{11}i}{6} \).
In simple words: Since the discriminant is negative, the roots are complex numbers. We use the quadratic formula to find them, replacing \( \sqrt{-11} \) with \( \sqrt{11}i \).

🎯 Exam Tip: Always write the final roots separately at the end of your solution to make it easy for the examiner to award full marks.

Question 2. Solve the quadratic equation:
(iv) \( x^2 - 4x + 13 = 0 \)
Answer:
Given equation is \( x^2 - 4x + 13 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get:
\( a = 1, b = -4, c = 13 \)
Discriminant \( = b^2 - 4ac \)
\( = (-4)^2 - 4 \times 1 \times 13 \)
\( = 16 - 52 \)
\( = -36 < 0 \)
So, the given equation has complex roots. This shows that the roots are complex conjugates of each other.
These roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ = \frac{-(-4) \pm \sqrt{-36}}{2(1)} \]
\[ = \frac{4 \pm 6i}{2} = 2 \pm 3i \]
\( \therefore \) the roots of the given equation are \( 2 + 3i \) and \( 2 - 3i \).
In simple words: Since the discriminant is negative, the equation has no real solutions, only complex ones. We find these by using the quadratic formula, which gives us two complex numbers as answers.

🎯 Exam Tip: When the discriminant \( b^2 - 4ac \) is negative, remember to write the square root of the negative number using the imaginary unit \( i \) (where \( \sqrt{-1} = i \)) to avoid calculation errors.

Question 3. Solve the following quadratic equations.
(i) \( x^2 + 3ix + 10 = 0 \)
Answer:
Given equation is \( x^2 + 3ix + 10 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get:
\( a = 1, b = 3i, c = 10 \)
Discriminant \( = b^2 - 4ac \)
\( = (3i)^2 - 4 \times 1 \times 10 \)
\( = 9i^2 - 40 \)
\( = -9 - 40 \)
\( = -49 < 0 \)
So, the given equation has complex roots. These imaginary roots satisfy the original quadratic equation perfectly.
These roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ = \frac{-3i \pm \sqrt{-49}}{2(1)} \]
\[ = \frac{-3i \pm 7i}{2} \]

\( \implies x = \frac{-3i + 7i}{2} = \frac{4i}{2} = 2i \)

\( \implies x = \frac{-3i - 7i}{2} = \frac{-10i}{2} = -5i \)
\( \therefore \) the roots of the given equation are \( 2i \) and \( -5i \).
In simple words: We solve this quadratic equation using the standard formula, keeping in mind that \( i^2 = -1 \). This gives us two purely imaginary numbers as the final roots.

🎯 Exam Tip: Be extremely careful when squaring terms with \( i \); remember that \( (3i)^2 = 9i^2 = -9 \). A common mistake is forgetting to change the sign.

Question (i) [Continuation of Solution]
Answer:
\( = (3i)^2 - 4 \times 1 \times 10 \)
\( = 9i^2 - 40 \)
\( = -9 - 40 \quad [\because i^2 = -1] \)
\( = -49 \)
So, the given equation has complex roots.
These roots are given by
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3i \pm \sqrt{-49}}{2(1)} \]
\[ x = \frac{-3i \pm 7i}{2} \]
\[ x = \frac{-3i + 7i}{2} \text{ or } x = \frac{-3i - 7i}{2} \]
\( \therefore x = 2i \) or \( x = -5i \)
\( \therefore \) the roots of the given equation are \( 2i \) and \( -5i \).

Check:
If \( x = 2i \) and \( x = -5i \) satisfy the given equation, then our answer is correct.
L.H.S. \( = x^2 + 3ix + 10 \)
\( = (2i)^2 + 3i(2i) + 10 \)
\( = 4i^2 + 6i^2 + 10 \)
\( = 10i^2 + 10 \)
\( = -10 + 10 \quad [\because i^2 = -1] \)
\( = 0 \)
\( = \) R.H.S.

L.H.S. \( = x^2 + 3ix + 10 \)
\( = (-5i)^2 + 3i(-5i) + 10 \)
\( = 25i^2 - 15i^2 + 10 \)
\( = 10i^2 + 10 \)
\( = -10 + 10 \quad [\text{since } i^2 = -1] \)
\( = 0 \)
\( = \) R.H.S.
Thus, our answer is correct.
In simple words: We find the roots of the quadratic equation using the formula, and then we plug these roots back into the original equation to verify that both sides equal zero.

🎯 Exam Tip: Always show the verification step (L.H.S. = R.H.S.) when asked, as it guarantees full marks and helps you double-check your calculations.

Question (ii) \( 2x^2 + 3ix + 2 = 0 \)
Answer:
Given equation is \( 2x^2 + 3ix + 2 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 2, b = 3i, c = 2 \)
Discriminant \( = b^2 - 4ac \)
\( = (3i)^2 - 4 \times 2 \times 2 \)
\( = 9i^2 - 16 \)
\( = -9 - 16 \quad [\because i^2 = -1] \)
\( = -25 \)
So, the given equation has complex roots.
These roots are given by
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3i \pm \sqrt{-25}}{2(2)} \]
\[ x = \frac{-3i \pm 5i}{4} \]
\[ x = \frac{-3i + 5i}{4} \text{ or } x = \frac{-3i - 5i}{4} \]
\( \therefore x = \frac{2i}{4} = \frac{i}{2} \) or \( x = \frac{-8i}{4} = -2i \)
\( \therefore \) the roots of the given equation are \( \frac{i}{2} \) and \( -2i \).
In simple words: We compare the equation with the standard quadratic form to find the coefficients, calculate the discriminant, and use the quadratic formula to find the complex roots.

🎯 Exam Tip: Remember that \( i^2 = -1 \). When simplifying the discriminant, always replace \( i^2 \) with \( -1 \) before adding or subtracting the terms.

Question 1. Solve the following quadratic equations:
(ii) \( 2x^2 + 3ix + 2 = 0 \)
(iii) \( x^2 + 4ix - 4 = 0 \)
(iv) \( ix^2 - 4x - 4i = 0 \)
Answer:
(ii) \( 2x^2 + 3ix + 2 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 2, b = 3i, c = 2 \).
Discriminant \( = b^2 - 4ac \)
\( = (3i)^2 - 4 \times 2 \times 2 \)
\( = 9i^2 - 16 \)
\( = -9 - 16 \quad [\because i^2 = -1] \)
\( = -25 < 0 \)
So, the given equation has complex roots. These roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ = \frac{-3i \pm \sqrt{-25}}{2(2)} \]
\[ x = \frac{-3i \pm 5i}{4} \]
\[ x = \frac{-3i + 5i}{4} \text{ or } x = \frac{-3i - 5i}{4} \]
\[ x = \frac{1}{2}i \text{ or } x = -2i \]
\( \dots \) the roots of the given equation are \( \frac{1}{2}i \) and \( -2i \).

(iii) \( x^2 + 4ix - 4 = 0 \)
Given equation is \( x^2 + 4ix - 4 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = 4i, c = -4 \)
Discriminant \( = b^2 - 4ac \)
\( = (4i)^2 - 4 \times 1 \times -4 \)
\( = 16i^2 + 16 \)
\( = -16 + 16 \quad [\because i^2 = -1] \)
\( = 0 \)
So, the given equation has equal roots. These roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
\[ = \frac{-4i \pm \sqrt{0}}{2(1)} = \frac{-4i}{2} \]
\[ x = -2i \]
\( \dots \) the roots of the given equation are \( -2i \) and \( -2i \).

(iv) \( ix^2 - 4x - 4i = 0 \)
Given equation is \( ix^2 - 4x - 4i = 0 \)
Multiplying throughout by \( i \), we get
\( i(ix^2 - 4x - 4i) = 0 \)

\( \implies i^2x^2 - 4ix - 4i^2 = 0 \)

\( \implies -x^2 - 4ix + 4 = 0 \quad [\because i^2 = -1] \)

\( \implies x^2 + 4ix - 4 = 0 \)
Comparing this simplified equation with the standard form \( ax^2 + bx + c = 0 \), we find that it is identical to sub-question (iii).
Therefore, the roots are given by:
\[ x = \frac{-4i \pm \sqrt{0}}{2(1)} \]
\[ x = -2i \]
\( \dots \) the roots of the given equation are \( -2i \) and \( -2i \).
In simple words: To solve quadratic equations containing imaginary numbers, we use the standard quadratic formula. If the term under the square root (discriminant) is negative, we use the imaginary unit \( i \) to write the complex roots, and if it is zero, the equation has two identical roots.

🎯 Exam Tip: When simplifying equations with \( i \), always remember that \( i^2 = -1 \). Double-check your signs when multiplying terms by \( i \) to avoid simple calculation errors.

Question 3. Solve the quadratic equation \(i^2x^2 - 4ix - 4i^2 = 0\).
Answer: Given equation is \(i^2x^2 - 4ix - 4i^2 = 0\).
\(\therefore -x^2 - 4ix + 4 = 0\) [since \(i^2 = -1\)]
\(\therefore x^2 + 4ix - 4 = 0\)
Comparing with \(ax^2 + bx + c = 0\), we get:
\(a = 1, b = 4i, c = -4\)
Discriminant = \(b^2 - 4ac\)
\(= (4i)^2 - 4 \times 1 \times (-4)\)
\(= 16i^2 + 16\)
\(= -16 + 16\) [since \(i^2 = -1\)]
\(= 0\)
So, the given equation has equal roots. Since the discriminant is zero, the quadratic equation yields two identical real or complex roots depending on the coefficients.
These roots are given by:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(= \frac{-4i \pm \sqrt{0}}{2(1)}\)
\(= \frac{-4i}{2}\)
\(= -2i\)
\(\therefore\) the roots of the given equation are \(-2i\) and \(-2i\).
In simple words: We simplify the equation using the fact that \(i^2 = -1\), and then use the quadratic formula to find that both roots are exactly the same.

🎯 Exam Tip: Always substitute \(i^2 = -1\) at the very beginning to simplify the quadratic equation before comparing coefficients.

Question 4. Solve the following quadratic equations.
(i) \(x^2 - (2 + i) x - (1 - 7i) = 0\)
Answer: Given equation is \(x^2 - (2 + i)x - (1 - 7i) = 0\).
Comparing with \(ax^2 + bx + c = 0\), we get:
\(a = 1, b = -(2 + i), c = -(1 - 7i)\)
Discriminant = \(b^2 - 4ac\)
\(= [-(2 + i)]^2 - 4 \times 1 \times [-(1 - 7i)]\)
\(= 4 + 4i + i^2 + 4 - 28i\)
\(= 4 + 4i - 1 + 4 - 28i\) [since \(i^2 = -1\)]
\(= 7 - 24i\)
So, the given equation has complex roots. The roots of a quadratic equation with complex coefficients do not necessarily occur in conjugate pairs, which is a key distinction from equations with real coefficients.
These roots are given by:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
\(x = \frac{(2 + i) \pm \sqrt{7 - 24i}}{2}\)
To find \(\sqrt{7 - 24i}\), let \(\sqrt{7 - 24i} = a + ib\), where \(a, b \in \mathbb{R}\).
Squaring both sides, we get:
\(7 - 24i = (a + ib)^2\)
\(7 - 24i = a^2 - b^2 + 2abi\)
Equating real and imaginary parts, we get:
\(a^2 - b^2 = 7\) --- (1)
\(2ab = -24\)
\( \implies \) \(ab = -12\) --- (2)
Now, \((a^2 + b^2)^2 = (a^2 - b^2)^2 + 4a^2b^2\)
\(= (7)^2 + 4(-12)^2\)
\(= 49 + 576\)
\(= 625\)
\(\therefore (a^2 + b^2)^2 = 625\)
\( \implies \) \(a^2 + b^2 = 25\) --- (3) [since \(a^2 + b^2 > 0\)]
Adding (1) and (3), we get:
\(2a^2 = 32\)
\( \implies \) \(a^2 = 16\)
\( \implies \) \(a = \pm 4\)
Subtracting (1) from (3), we get:
\(2b^2 = 18\)
\( \implies \) \(b^2 = 9\)
\( \implies \) \(b = \pm 3\)
Since \(ab = -12\) is negative, \(a\) and \(b\) must have opposite signs.
Therefore, \(\sqrt{7 - 24i} = \pm(4 - 3i)\).
Substituting this value back into the formula for \(x\):
\(x = \frac{(2 + i) \pm (4 - 3i)}{2}\)
Taking the positive sign:
\(x = \frac{2 + i + 4 - 3i}{2} = \frac{6 - 2i}{2} = 3 - i\)
Taking the negative sign:
\(x = \frac{2 + i - (4 - 3i)}{2} = \frac{2 + i - 4 + 3i}{2} = \frac{-2 + 4i}{2} = -1 + 2i\)
Thus, the roots of the given quadratic equation are \(3 - i\) and \(-1 + 2i\).
In simple words: To solve this quadratic equation, we use the standard formula just like with regular numbers, but since the discriminant is a complex number, we find its square root by setting up a system of equations for its real and imaginary parts.

🎯 Exam Tip: When finding the square root of a complex number \(a + ib\), always remember that the sign of the product \(xy\) determines whether \(x\) and \(y\) have the same or opposite signs.

Question 1. Solve the following quadratic equations:
(i) \( x^2 - (2+i)x + (-1+7i) = 0 \)
(ii) \( x^2 - (3\sqrt{2} + 2i)x + 6\sqrt{2}i = 0 \)
Answer:
(i) Given equation is \( x^2 - (2+i)x + (-1+7i) = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = -(2+i), c = -1+7i \)
Using the quadratic formula:
\( \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \implies x = \frac{-[-(2+i)] \pm \sqrt{[-(2+i)]^2 - 4(1)(-1+7i)}}{2(1)} \)
\( \implies x = \frac{(2+i) \pm \sqrt{7-24i}}{2} \)
Let \( \sqrt{7-24i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( 7 - 24i = a^2 + i^2b^2 + 2abi \)
\( \implies 7 - 24i = (a^2 - b^2) + 2abi \) ...[\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = 7 \) and \( 2ab = -24 \)
\( \implies a^2 - b^2 = 7 \) and \( b = \frac{-12}{a} \)
Substituting the value of \( b \) in the first equation:
\( \implies a^2 - \left(\frac{-12}{a}\right)^2 = 7 \)
\( \implies a^2 - \frac{144}{a^2} = 7 \)
\( \implies a^4 - 144 = 7a^2 \)
\( \implies a^4 - 7a^2 - 144 = 0 \)
\( \implies (a^2 - 16)(a^2 + 9) = 0 \)
\( \implies a^2 = 16 \) or \( a^2 = -9 \)
But \( a \in \mathbb{R} \), so \( a^2 \) cannot be negative.
\( \implies a^2 \neq -9 \)
\( \implies a^2 = 16 \)
\( \implies a = \pm 4 \)
When \( a = 4 \), \( b = \frac{-12}{4} = -3 \)
When \( a = -4 \), \( b = \frac{-12}{-4} = 3 \)
\( \implies \sqrt{7-24i} = \pm(4 - 3i) \)
Substituting this back into the expression for \( x \):
\( \implies x = \frac{(2+i) \pm (4-3i)}{2} \)
\( \implies x = \frac{(2+i) + (4-3i)}{2} \) or \( x = \frac{(2+i) - (4-3i)}{2} \)
\( \implies x = 3 - i \) or \( x = -1 + 2i \)

(ii) Given equation is \( x^2 - (3\sqrt{2} + 2i)x + 6\sqrt{2}i = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = -(3\sqrt{2} + 2i), c = 6\sqrt{2}i \)
Using the quadratic formula:
\( \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \implies x = \frac{-[-(3\sqrt{2} + 2i)] \pm \sqrt{[-(3\sqrt{2} + 2i)]^2 - 4(1)(6\sqrt{2}i)}}{2(1)} \)
\( \implies x = \frac{(3\sqrt{2} + 2i) \pm \sqrt{(18 - 4 + 12\sqrt{2}i) - 24\sqrt{2}i}}{2} \)
\( \implies x = \frac{(3\sqrt{2} + 2i) \pm \sqrt{14 - 12\sqrt{2}i}}{2} \)
Notice that \( 14 - 12\sqrt{2}i = (3\sqrt{2})^2 + (2i)^2 - 2(3\sqrt{2})(2i) = (3\sqrt{2} - 2i)^2 \).
\( \implies x = \frac{(3\sqrt{2} + 2i) \pm (3\sqrt{2} - 2i)}{2} \)
\( \implies x = \frac{(3\sqrt{2} + 2i) + (3\sqrt{2} - 2i)}{2} \) or \( x = \frac{(3\sqrt{2} + 2i) - (3\sqrt{2} - 2i)}{2} \)
\( \implies x = \frac{6\sqrt{2}}{2} \) or \( x = \frac{4i}{2} \)
\( \implies x = 3\sqrt{2} \) or \( x = 2i \)
In simple words: To solve these quadratic equations with complex numbers, we use the standard quadratic formula. When finding the square root of a complex number, we assume it equals \( a + bi \), square both sides, and solve for the real numbers \( a \) and \( b \) to get the final roots.

🎯 Exam Tip: When finding the square root of a complex number under the radical, always remember that \( a \) and \( b \) must be real numbers, so discard any negative values for \( a^2 \).

Question 1. Solve the quadratic equation to find its complex roots.
Answer: To find the roots of the quadratic equation, we first calculate the discriminant:
Discriminant = \( b^2 - 4ac \)
\( = [-(3\sqrt{2} + 2i)]^2 - 4 \times 1 \times 6\sqrt{2}i \)
\( = 18 + 12\sqrt{2}i + 4i^2 - 24\sqrt{2}i \)
\( = 18 - 12\sqrt{2}i - 4 \quad [\because i^2 = -1] \)
\( = 14 - 12\sqrt{2}i \)
So, the given equation has complex roots. These roots are given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-[-(3\sqrt{2} + 2i)] \pm \sqrt{14 - 12\sqrt{2}i}}{2(1)} \)
\( = \frac{(3\sqrt{2} + 2i) \pm \sqrt{14 - 12\sqrt{2}i}}{2} \)
Let \( \sqrt{14 - 12\sqrt{2}i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( 14 - 12\sqrt{2}i = a^2 + i^2b^2 + 2abi \)

\( \implies 14 - 12\sqrt{2}i = (a^2 - b^2) + 2abi \quad [\because i^2 = -1] \)
In simple words: To solve a quadratic equation with complex numbers, we find the discriminant first. Since the discriminant contains an imaginary part, we find its square root by setting it equal to \( a + bi \) and squaring both sides to compare the real and imaginary parts.

🎯 Exam Tip: When squaring a complex number like \( a + bi \), always remember that \( i^2 = -1 \), which changes the sign of the \( b^2 \) term to negative.

Question. Solve the quadratic equation (continued):
Answer: Equating real and imaginary parts, we get:
\( a^2 - b^2 = 14 \) and \( 2ab = -12\sqrt{2} \)

\( \implies a^2 - b^2 = 14 \) and \( b = \frac{-6\sqrt{2}}{a} \)

\( \implies a^2 - \left(\frac{-6\sqrt{2}}{a}\right)^2 = 14 \)

\( \implies a^2 - \frac{72}{a^2} = 14 \)

\( \implies a^4 - 72 = 14a^2 \)

\( \implies a^4 - 14a^2 - 72 = 0 \)

\( \implies (a^2 - 18)(a^2 + 4) = 0 \)

\( \implies a^2 = 18 \) or \( a^2 = -4 \)
But \( a \in \mathbb{R} \)

\( \implies a^2 \neq -4 \)

\( \implies a^2 = 18 \)

\( \implies a = \pm 3\sqrt{2} \)

When \( a = 3\sqrt{2} \), \( b = \frac{-6\sqrt{2}}{3\sqrt{2}} = -2 \)

When \( a = -3\sqrt{2} \), \( b = \frac{-6\sqrt{2}}{-3\sqrt{2}} = 2 \)

\( \implies \sqrt{14-12\sqrt{2}i} = \pm \left(3\sqrt{2}-2i\right) \)

\( \implies x = \frac{(3\sqrt{2} + 2i) \pm (3\sqrt{2} - 2i)}{2} \)

\( \implies x = \frac{(3\sqrt{2} + 2i) + (3\sqrt{2} - 2i)}{2} \) or \( x = \frac{(3\sqrt{2} + 2i) - (3\sqrt{2} - 2i)}{2} \)

\( \implies x = 3\sqrt{2} \) or \( x = 2i \)
In simple words: We find the square root of the complex number by setting up equations for its real and imaginary parts, solving for \(a\) and \(b\), and then substituting these values back into the quadratic formula to find the final roots.

🎯 Exam Tip: Remember that since \( a \) is a real number (\( a \in \mathbb{R} \)), \( a^2 \) cannot be negative. Always state this reason clearly to discard the negative value of \( a^2 \).

Question. Solve the following quadratic equation:
(iii) \( x^2 - (5 - i)x + (18 + i) = 0 \)
Answer: Given equation is \( x^2 - (5 - i)x + (18 + i) = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = -(5 - i), c = 18 + i \)
Discriminant = \( b^2 - 4ac \)
\( = [-(5 - i)]^2 - 4 \times 1 \times (18 + i) \)
\( = 25 - 10i + i^2 - 72 - 4i \)
\( = 25 - 10i - 1 - 72 - 4i \quad [\because i^2 = -1] \)
\( = -48 - 14i \)
So, the given equation has complex roots. These roots are given by the quadratic formula.
In simple words: We compare the given equation with the standard quadratic form to find the coefficients, and then calculate the discriminant to determine the roots.

🎯 Exam Tip: When finding the discriminant of a quadratic equation with complex coefficients, be extremely careful with the signs and remember that \( i^2 = -1 \).

Question 1. Solve the quadratic equation:
(iii) \( x^2 - (5 - i)x + (18 + i) = 0 \)
Answer: Given equation is \( x^2 - (5 - i)x + (18 + i) = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 1, b = -(5 - i), c = 18 + i \)
Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( \implies x = \frac{-[-(5-i)] \pm \sqrt{[-(5-i)]^2 - 4(1)(18+i)}}{2(1)} \)
\( \implies x = \frac{(5-i) \pm \sqrt{-48-14i}}{2} \)
Let \( \sqrt{-48-14i} = a + bi \), where \( a, b \in \mathbb{R} \)
Squaring on both sides, we get
\( -48 - 14i = a^2 + b^2i^2 + 2abi \)
\( -48 - 14i = (a^2 - b^2) + 2abi \) ...[\( \because i^2 = -1 \)]
Equating real and imaginary parts, we get
\( a^2 - b^2 = -48 \) and \( 2ab = -14 \)
\( \therefore a^2 - b^2 = -48 \) and \( b = \frac{-7}{a} \)
\( \therefore a^2 - \left(\frac{-7}{a}\right)^2 = -48 \)
\( \dots a^2 - \frac{49}{a^2} = -48 \)
\( \therefore a^4 - 49 = -48a^2 \)
\( \therefore a^4 + 48a^2 - 49 = 0 \)
\( \therefore (a^2 + 49)(a^2 - 1) = 0 \)
\( \therefore a^2 = -49 \) or \( a^2 = 1 \)
But \( a \in \mathbb{R} \)
\( \therefore a^2 \neq -49 \)
\( \therefore a^2 = 1 \)
\( \therefore a = \pm 1 \)
When \( a = 1 \), \( b = \frac{-7}{1} = -7 \)
When \( a = -1 \), \( b = \frac{-7}{-1} = 7 \)
\( \therefore \sqrt{-48-14i} = \pm(1 - 7i) \)
\( \therefore x = \frac{(5-i) \pm (1-7i)}{2} \)
\( \therefore x = \frac{5-i+1-7i}{2} \) or \( x = \frac{5-i-1+7i}{2} \)
\( \therefore x = 3 - 4i \) or \( x = 2 + 3i \). This step-by-step algebraic reduction ensures we find all valid complex roots.
In simple words: To solve this quadratic equation with complex numbers, we use the standard quadratic formula and then find the square root of the complex number under the radical by setting it equal to \( a + bi \).

🎯 Exam Tip: When finding the square root of a complex number \( a + bi \), remember that \( a \) and \( b \) must be real numbers, so discard any negative values for \( a^2 \).

Question 1. Solve the quadratic equation:
(iv) \( (2 + i)x^2 - (5 - i)x + 2(1 - i) = 0 \)
Answer: Given equation is \( (2 + i)x^2 - (5 - i)x + 2(1 - i) = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get
\( a = 2 + i, b = -(5 - i), c = 2(1 - i) \)
Using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, let's find the discriminant \( b^2 - 4ac \):
\( b^2 - 4ac = [-(5-i)]^2 - 4(2+i)[2(1-i)] \)
\( \implies b^2 - 4ac = (24 - 10i) - 8(2 - 2i + i + 1) \)
\( \implies b^2 - 4ac = (24 - 10i) - 8(3 - i) \)
\( \implies b^2 - 4ac = 24 - 10i - 24 + 8i \)
\( \implies b^2 - 4ac = -2i \)
Let \( \sqrt{-2i} = x_1 + y_1i \), where \( x_1, y_1 \in \mathbb{R} \)
Squaring both sides:
\( -2i = (x_1^2 - y_1^2) + 2x_1y_1i \)
Equating real and imaginary parts:
\( x_1^2 - y_1^2 = 0 \) and \( 2x_1y_1 = -2 \)
\( \implies y_1 = -\frac{1}{x_1} \)
\( \therefore x_1^2 - \left(-\frac{1}{x_1}\right)^2 = 0 \)
\( \dots x_1^4 - 1 = 0 \)
\( \therefore (x_1^2 - 1)(x_1^2 + 1) = 0 \)
Since \( x_1 \in \mathbb{R} \), \( x_1^2 = 1 \)
\( \implies x_1 = \pm 1 \)
When \( x_1 = 1 \), \( y_1 = -1 \)
When \( x_1 = -1 \), \( y_1 = 1 \)
\( \dots \sqrt{-2i} = \pm(1 - i) \)
Now, substitute this back into the quadratic formula:
\( x = \frac{(5-i) \pm (1-i)}{2(2+i)} \)
Either \( x = \frac{(5-i) + (1-i)}{2(2+i)} = \frac{6-2i}{2(2+i)} = \frac{3-i}{2+i} \)
\( \implies x = \frac{(3-i)(2-i)}{(2+i)(2-i)} = \frac{6 - 3i - 2i - 1}{4 + 1} = \frac{5 - 5i}{5} = 1 - i \)
Or \( x = \frac{(5-i) - (1-i)}{2(2+i)} = \frac{4}{2(2+i)} = \frac{2}{2+i} \)
\( \implies x = \frac{2(2-i)}{(2+i)(2-i)} = \frac{4 - 2i}{4 + 1} = \frac{4}{5} - \frac{2}{5}i \)
\( \therefore x = 1 - i \) or \( x = \frac{4}{5} - \frac{2}{5}i \). Carefully tracking the signs during complex multiplication prevents common arithmetic errors.
In simple words: We find the discriminant of the quadratic equation, simplify the square root of the resulting complex number, and then solve for \( x \) by rationalizing the denominator.

🎯 Exam Tip: When dividing by a complex number like \( 2+i \), always multiply the numerator and denominator by its conjugate \( 2-i \) to make the denominator a real number.

Question 1. Solve the quadratic equation \( (2 + i)x^2 - (5 - i)x + 2(1 - i) = 0 \).
Answer:
Given the quadratic equation, we identify the coefficients as:
\( a = 2 + i \)
\( b = -(5 - i) \)
\( c = 2(1 - i) \)

First, we find the discriminant:
\( \text{Discriminant} = b^2 - 4ac \)
\( = [-(5 - i)]^2 - 4 \times (2 + i) \times 2(1 - i) \)
\( = 25 - 10i + i^2 - 8(2 + i)(1 - i) \)
\( = 25 - 10i + i^2 - 8(2 - 2i + i - i^2) \)
\( = 25 - 10i - 1 - 8(2 - i + 1) \quad \dots [\because i^2 = -1] \)
\( = 25 - 10i - 1 - 16 + 8i - 8 \)
\( = -2i \)

So, the given equation has complex roots. These roots are given by:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( = \frac{-[-(5 - i)] \pm \sqrt{-2i}}{2(2 + i)} = \frac{(5 - i) \pm \sqrt{-2i}}{2(2 + i)} \)

To simplify this further, we need to find the square root of the complex number \( -2i \).
Let \( \sqrt{-2i} = a + bi \), where \( a, b \in \mathbb{R} \).
Squaring on both sides, we get:
\( -2i = a^2 + b^2i^2 + 2abi \)
\( 0 - 2i = (a^2 - b^2) + 2abi \quad \dots [\because i^2 = -1] \)

Equating real and imaginary parts, we get:
\( a^2 - b^2 = 0 \) and \( 2ab = -2 \)
\( a^2 - b^2 = 0 \) and \( b = -\frac{1}{a} \)
This system of equations can be solved to find the real values of \( a \) and \( b \), which will then give us the square root of the complex number.
In simple words: To solve a quadratic equation with complex numbers, we first calculate the discriminant just like we do with normal numbers. Since the discriminant is a complex number, we find its square root by setting it equal to \( a + bi \) and solving for the real numbers \( a \) and \( b \).

🎯 Exam Tip: When finding the square root of a complex number \( x + iy \), always equate real and imaginary parts carefully and remember that \( a \) and \( b \) must be real numbers.

Question. Solve the complex equation to find the values of \( a \), \( b \), and \( x \).
Answer: We begin by solving the equation for the real variable \( a \):
\( a^2 - \left(-\frac{1}{a}\right)^2 = 0 \)
\( \implies a^2 - \frac{1}{a^2} = 0 \)
\( \implies a^4 - 1 = 0 \)
\( \implies (a^2 - 1)(a^2 + 1) = 0 \)
\( \implies a^2 = 1 \) or \( a^2 = -1 \)
But \( a \in \mathbb{R} \)
\( \implies a^2 \neq -1 \)
\( \implies a^2 = 1 \)
\( \implies a = \pm 1 \)
When \( a = 1 \), \( b = -1 \)
When \( a = -1 \), \( b = 1 \)
\( \implies \sqrt{-2i} = \pm(1 - i) \)
Now, substituting these values to solve for \( x \):
\( \implies x = \frac{(5-i) \pm (1-i)}{2(2+i)} \)
\( \implies x = \frac{5-i+1-i}{2(2+i)} \) or \( x = \frac{5-i-1+i}{2(2+i)} \)
\( \implies x = \frac{6-2i}{2(2+i)} \) or \( x = \frac{4}{2(2+i)} \)
\( \implies x = \frac{2(3-i)}{2(2+i)} \) or \( x = \frac{2}{2+i} \)
\( \implies x = \frac{3-i}{2+i} \) or \( x = \frac{2(2-i)}{(2+i)(2-i)} \)
\( \implies x = \frac{(3-i)(2-i)}{(2+i)(2-i)} \) or \( x = \frac{2(2-i)}{4-i^2} \)
\( \implies x = \frac{6-5i+i^2}{4-i^2} \) or \( x = \frac{4-2i}{4-i^2} \)
\( \implies x = \frac{5-5i}{5} \) or \( x = \frac{4-2i}{5} \quad [\because i^2 = -1] \)
\( \implies x = 1-i \) or \( x = \frac{4}{5} - \frac{2i}{5} \)
In simple words: First, we find the real values of \( a \) and \( b \) by solving the algebraic equation. Then, we substitute these values back into the expression for \( x \) and simplify the complex fractions by multiplying the numerator and denominator by the conjugate of the denominator.

🎯 Exam Tip: When simplifying complex fractions, always multiply the numerator and denominator by the conjugate of the denominator to eliminate the imaginary unit \( i \) from the denominator. Remember that \( i^2 = -1 \).