RBSE Solutions Class 11 Maths Chapter 5 Complex Numbers Exercise 5.4

Question 1. Find the solution of following equations by Vedic method
(i) \( x^2 + 4x + 13 = 0 \)
(ii) \( 2x^2 + 5x + 4 = 0 \)
(iii) \( ix^2 + 4x - \frac{15}{2} = 0 \)
Answer:
(i) We are given the equation \( x^2 + 4x + 13 = 0 \). Comparing this to the standard quadratic equation \( ax^2 + bx + c = 0 \), we identify \( a = 1 \), \( b = 4 \), and \( c = 13 \). Using the Vedic method, which involves the formula \( 2ax + b = \pm\sqrt{b^2 - 4ac} \), we substitute the values: \( 2(1)x + 4 = \pm\sqrt{4^2 - 4(1)(13)} \) \( 2x + 4 = \pm\sqrt{16 - 52} \) \( 2x + 4 = \pm\sqrt{-36} \) Since \( \sqrt{-36} = 6i \): \( 2x + 4 = \pm 6i \) Subtract 4 from both sides: \( 2x = -4 \pm 6i \) Divide by 2 to find x: \( x = \frac{-4 \pm 6i}{2} \) \( x = -2 \pm 3i \) So, the solutions for the equation \( x^2 + 4x + 13 = 0 \) are \( -2 + 3i \) and \( -2 - 3i \).
(ii) We are given the equation \( 2x^2 + 5x + 4 = 0 \). Comparing this with \( ax^2 + bx + c = 0 \), we find that \( a = 2 \), \( b = 5 \), and \( c = 4 \). Applying the Vedic method formula \( 2ax + b = \pm\sqrt{b^2 - 4ac} \): \( 2(2)x + 5 = \pm\sqrt{5^2 - 4(2)(4)} \) \( 4x + 5 = \pm\sqrt{25 - 32} \) \( 4x + 5 = \pm\sqrt{-7} \) Since \( \sqrt{-7} = i\sqrt{7} \): \( 4x + 5 = \pm i\sqrt{7} \) Subtract 5 from both sides: \( 4x = -5 \pm i\sqrt{7} \) Divide by 4 to find x: \( x = \frac{-5 \pm i\sqrt{7}}{4} \) Thus, the solutions for the equation \( 2x^2 + 5x + 4 = 0 \) are \( \frac{-5 + i\sqrt{7}}{4} \) and \( \frac{-5 - i\sqrt{7}}{4} \).
(iii) The given equation is \( ix^2 + 4x - \frac{15}{2} = 0 \). To remove the fraction, multiply the entire equation by 2: \( 2ix^2 + 8x - 15 = 0 \) Comparing this with \( ax^2 + bx + c = 0 \), we identify \( a = 2i \), \( b = 8 \), and \( c = -15 \). Using the Vedic method formula \( 2ax + b = \pm\sqrt{b^2 - 4ac} \): \( 2(2i)x + 8 = \pm\sqrt{8^2 - 4(2i)(-15)} \) \( 4ix + 8 = \pm\sqrt{64 + 120i} \) To simplify \( \sqrt{64 + 120i} \), we note that \( \sqrt{16+30i} = \pm (5+3i) \). Since \( 64 + 120i = 4(16 + 30i) \), we can say: \( \sqrt{64 + 120i} = \sqrt{4(16 + 30i)} = 2\sqrt{16 + 30i} = 2 \times \pm (5+3i) = \pm (10+6i) \) So, \( 4ix + 8 = \pm (10+6i) \) Dividing the entire equation by 2: \( 2ix + 4 = \pm (5+3i) \) Subtract 4 from both sides: \( 2ix = -4 \pm (5+3i) \) Now, we find the two solutions for x: **Case 1:** \( 2ix = -4 + (5+3i) \) \( 2ix = 1 + 3i \) Divide by \( 2i \): \( x = \frac{1+3i}{2i} \) To simplify, multiply the numerator and denominator by \( i \): \( x = \frac{(1+3i)i}{2i \cdot i} = \frac{i + 3i^2}{2i^2} = \frac{i - 3}{-2} = \frac{3 - i}{2} \) **Case 2:** \( 2ix = -4 - (5+3i) \) \( 2ix = -4 - 5 - 3i \) \( 2ix = -9 - 3i \) Divide by \( 2i \): \( x = \frac{-9-3i}{2i} \) To simplify, multiply the numerator and denominator by \( i \): \( x = \frac{(-9-3i)i}{2i \cdot i} = \frac{-9i - 3i^2}{2i^2} = \frac{-9i + 3}{-2} = \frac{3 - 9i}{2} \) Thus, the solutions for the equation \( ix^2 + 4x - \frac{15}{2} = 0 \) are \( \frac{3 - i}{2} \) and \( \frac{3 - 9i}{2} \).
In simple words: To find the solutions for these equations, we use a special method called the Vedic method, which is similar to the quadratic formula. We identify the numbers a, b, and c from the equation and plug them into the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This helps us find the complex numbers that satisfy the equations.

🎯 Exam Tip: When dealing with complex numbers and quadratic equations, remember that complex roots always appear in conjugate pairs if the coefficients are real. Pay close attention to calculations involving \( i^2 = -1 \).

Question 2. Find quadratic equation which has the following roots
(i) 5 and -2
(ii) 1 + 2i
Answer:
(i) We are given the roots as \( \alpha = 5 \) and \( \beta = -2 \). First, calculate the sum of the roots: Sum \( = \alpha + \beta = 5 + (-2) = 3 \) Next, calculate the product of the roots: Product \( = \alpha\beta = 5 \times (-2) = -10 \) The general form of a quadratic equation when its roots are known is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \). Substitute the calculated sum and product into this formula: \( x^2 - (3)x + (-10) = 0 \) \( x^2 - 3x - 10 = 0 \) Therefore, the quadratic equation with roots 5 and -2 is \( x^2 - 3x - 10 = 0 \).
(ii) We are given one root as \( 1 + 2i \). For quadratic equations with real coefficients, complex roots always appear in conjugate pairs. So, if one root is \( \alpha = 1 + 2i \), then the other root must be its conjugate, \( \beta = 1 - 2i \). First, calculate the sum of the roots: Sum \( = \alpha + \beta = (1 + 2i) + (1 - 2i) = 1 + 1 + 2i - 2i = 2 \) Next, calculate the product of the roots: Product \( = \alpha\beta = (1 + 2i)(1 - 2i) \) Using the identity \( (a+b)(a-b) = a^2 - b^2 \): Product \( = 1^2 - (2i)^2 = 1 - 4i^2 \) Since \( i^2 = -1 \): Product \( = 1 - 4(-1) = 1 + 4 = 5 \) The general form of a quadratic equation is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \). Substitute the calculated sum and product: \( x^2 - (2)x + (5) = 0 \) \( x^2 - 2x + 5 = 0 \) Therefore, the quadratic equation with roots \( 1 + 2i \) and \( 1 - 2i \) is \( x^2 - 2x + 5 = 0 \).
In simple words: To create a quadratic equation from its roots, first add the roots together, then multiply them together. Once you have the sum and product, use the formula \( x^2 - (\text{sum})x + (\text{product}) = 0 \). Remember that if one root is a complex number, its partner root will be its conjugate.

🎯 Exam Tip: Always remember the fundamental relationships: Sum of roots \( = -\frac{b}{a} \) and Product of roots \( = \frac{c}{a} \). This is key to forming quadratic equations from given roots.

Question 4. Find that condition for which equation \( ax^2 + bx + c = 0 \) has roots in the ratio m : n.
Answer: We are given a quadratic equation \( ax^2 + bx + c = 0 \). Let its roots be \( \alpha \) and \( \beta \). According to the question, the roots are in the ratio \( m:n \). This means we can write \( \frac{\alpha}{\beta} = \frac{m}{n} \). From this ratio, we can express \( \alpha \) in terms of \( \beta \): \( \alpha = \frac{m}{n}\beta \). For a quadratic equation, we know two important relationships between roots and coefficients: 1. **Sum of roots:** \( \alpha + \beta = -\frac{b}{a} \) Substitute the expression for \( \alpha \): \( \frac{m}{n}\beta + \beta = -\frac{b}{a} \) Factor out \( \beta \): \( \beta \left( \frac{m}{n} + 1 \right) = -\frac{b}{a} \) Combine the terms inside the parenthesis: \( \beta \left( \frac{m+n}{n} \right) = -\frac{b}{a} \) Now, solve for \( \beta \): \( \beta = -\frac{bn}{a(m+n)} \) 2. **Product of roots:** \( \alpha\beta = \frac{c}{a} \) Substitute the expression for \( \alpha \): \( \left( \frac{m}{n}\beta \right) \beta = \frac{c}{a} \) \( \frac{m}{n}\beta^2 = \frac{c}{a} \) Now, solve for \( \beta^2 \): \( \beta^2 = \frac{cn}{am} \) Now we equate the square of the expression for \( \beta \) from the sum of roots to the expression for \( \beta^2 \) from the product of roots: \( \left( -\frac{bn}{a(m+n)} \right)^2 = \frac{cn}{am} \) Square the left side: \( \frac{b^2n^2}{a^2(m+n)^2} = \frac{cn}{am} \) To simplify, we can cancel one \( n \) from both sides (assuming \( n \ne 0 \)) and one \( a \) from the denominator of both sides (assuming \( a \ne 0 \)): \( \frac{b^2n}{a(m+n)^2} = \frac{c}{m} \) Now, cross-multiply: \( m b^2 n = c a (m+n)^2 \) Rearranging, the required condition is \( mnb^2 = ac(m+n)^2 \).
In simple words: If the roots of a quadratic equation are in a certain ratio, we can use the sum and product of roots formulas to find a relationship between the equation's coefficients (a, b, c) and that ratio (m:n). This relationship, \( mnb^2 = ac(m+n)^2 \), is the condition we are looking for.

🎯 Exam Tip: When given a ratio of roots, express one root in terms of the other and the ratio. Then, apply the sum and product of roots formulas to establish relationships and solve for the required condition. Be careful with algebraic manipulation.