RD Sharma Solutions for Class 11 Mathematics Chapter 15: Linear Inequations

RD Sharma Solutions for Class 11 Mathematics Chapter 15, “Linear Inequations,” are available here, solved by expert teachers following the latest CBSE syllabus and NCERT guidelines. On this page, students will find a wide variety of linear inequations problems with detailed step-by-step answers. RD Sharma Solution for Linear Inequations help students grasp the concept of inequalities and effectively solve linear inequations involving one and two variables. Students will also learn how to represent solutions graphically—an essential skill for mastering this chapter and scoring higher marks in exams.

Topics Covered in RD Sharma Sharma Solutions for Linear Inequations:

• Algebraic solutions for linear inequalities in one variable
• Graphical representation of solutions on a number line and coordinate plane
• Systems of linear inequations in two variables and their solutions
• Practice problems with detailed explanations to reinforce concepts

This chapter builds on your knowledge of equations and introduces inequalities, enabling you to analyze and solve problems both algebraically and graphically with confidence.

Download RD Sharma Class 11 Chapter 15 Linear Inequations Solutions PDF Free

Get comprehensive and exam-oriented RD Sharma Class 11 Maths solutions for Chapter 15, including solved examples, extra practice questions, and step-by-step explanations. Learn how to solve linear inequations in one variable, represent solutions on a number line, and solve systems of linear inequalities graphically.

Download the free PDF to boost your Class 11 Maths preparation with trusted RD Sharma solutions designed to help you master linear inequalities and excel in your exams.

RD Sharma Solutions for Linear Inequations: Solved Examples

Question 1: Express the equation 2x² - 5x + 3 = 0 in standard form and identify the coefficients.

Solution:

Step 1: The given equation is already in standard form: ax² + bx + c = 0

Step 2: Comparing 2x² - 5x + 3 = 0 with ax² + bx + c = 0

Answer: a = 2, b = -5, c = 3

Question 2: Solve the quadratic equation x² - 7x + 12 = 0 by factorization.

Solution:

Step 1: Find two numbers that multiply to 12 and add to -7

Step 2: The numbers are -3 and -4 (since -3 × -4 = 12 and -3 + (-4) = -7)

Step 3: Factor: x² - 7x + 12 = (x - 3)(x - 4) = 0

Step 4: Set each factor to zero: x - 3 = 0 or x - 4 = 0

Answer: x = 3 or x = 4

Question 3: Find the roots of the equation 3x² + 2x - 1 = 0 using the quadratic formula.

Solution:

Step 1: Identify coefficients: a = 3, b = 2, c = -1

Step 2: Apply quadratic formula: x = (-b ± √(b² - 4ac)) / (2a)

Step 3: Calculate discriminant: b² - 4ac = 4 - 4(3)(-1) = 4 + 12 = 16

Step 4: x = (-2 ± √16) / (2×3) = (-2 ± 4) / 6

Step 5: x = (-2 + 4) / 6 = 2/6 = 1/3 or x = (-2 - 4) / 6 = -6/6 = -1

Answer: x = 1/3 or x = -1

Question 4: Determine the nature of roots for the equation x² + 4x + 5 = 0.

Solution:

Step 1: Identify coefficients: a = 1, b = 4, c = 5

Step 2: Calculate discriminant: Δ = b² - 4ac = 16 - 4(1)(5) = 16 - 20 = -4

Step 3: Since Δ < 0, the roots are complex (imaginary)

Answer: The roots are complex and unequal

Question 5: If α and β are the roots of x² - 6x + 8 = 0, find the value of α + β and αβ.

Solution:

Step 1: For equation ax² + bx + c = 0, sum of roots = -b/a and product of roots = c/a

Step 2: Here a = 1, b = -6, c = 8

Step 3: α + β = -(-6)/1 = 6

Step 4: αβ = 8/1 = 8

Answer: α + β = 6, αβ = 8

Question 6: Solve the equation 2x² - 3x + 7 = 0 by completing the square method.

Solution:

Step 1: Divide by coefficient of x²: x² - (3/2)x + 7/2 = 0

Step 2: Move constant to right: x² - (3/2)x = -7/2

Step 3: Complete the square: x² - (3/2)x + (3/4)² = -7/2 + 9/16

Step 4: (x - 3/4)² = -56/16 + 9/16 = -47/16

Step 5: Since right side is negative, roots are complex

Step 6: x - 3/4 = ±i√(47/16) = ±i√47/4

Answer: x = 3/4 ± i√47/4

Question 7: Form a quadratic equation whose roots are 2 and -3.

Solution:

Step 1: If roots are α and β, then equation is (x - α)(x - β) = 0

Step 2: With roots 2 and -3: (x - 2)(x + 3) = 0

Step 3: Expand: x² + 3x - 2x - 6 = 0

Answer: x² + x - 6 = 0

Question 8: If one root of the equation x² + px + 12 = 0 is 3, find the value of p and the other root.

Solution:

Step 1: Since 3 is a root, substitute x = 3: 9 + 3p + 12 = 0

Step 2: 21 + 3p = 0, so p = -7

Step 3: For equation x² - 7x + 12 = 0, product of roots = 12

Step 4: If one root is 3, other root = 12/3 = 4

Answer: p = -7, other root = 4

Question 9: Find the quadratic equation whose roots are reciprocal of the roots of x² - 5x + 6 = 0.

Solution:

Step 1: First find roots of x² - 5x + 6 = 0

Step 2: Factor: (x - 2)(x - 3) = 0, so roots are 2 and 3

Step 3: Reciprocals are 1/2 and 1/3

Step 4: Sum of new roots = 1/2 + 1/3 = 5/6

Step 5: Product of new roots = (1/2)(1/3) = 1/6

Step 6: Required equation: x² - (5/6)x + 1/6 = 0

Step 7: Multiply by 6: 6x² - 5x + 1 = 0

Answer: 6x² - 5x + 1 = 0

Question 10: Solve for x: 4x² + 4x + 1 = 0.

Solution:

Step 1: Notice this is a perfect square: (2x + 1)² = 0

Step 2: Therefore 2x + 1 = 0

Step 3: x = -1/2

Answer: x = -1/2 (repeated root)

Question 11: If the roots of ax² + bx + c = 0 are equal, show that b² = 4ac.

Solution:

Step 1: For equal roots, discriminant = 0

Step 2: Discriminant = b² - 4ac

Step 3: Setting discriminant = 0: b² - 4ac = 0

Step 4: Therefore b² = 4ac

Answer: Hence proved

Question 12: Find the sum and product of the roots of 5x² - 2x + 1 = 0.

Solution:

Step 1: For ax² + bx + c = 0, sum = -b/a, product = c/a

Step 2: Here a = 5, b = -2, c = 1

Step 3: Sum of roots = -(-2)/5 = 2/5

Step 4: Product of roots = 1/5

Answer: Sum = 2/5, Product = 1/5

Question 13: Solve the quadratic equation x² - 2√2x + 2 = 0.

Solution:

Step 1: Use quadratic formula with a = 1, b = -2√2, c = 2

Step 2: Discriminant = (-2√2)² - 4(1)(2) = 8 - 8 = 0

Step 3: Since discriminant = 0, roots are equal

Step 4: x = 2√2/(2×1) = √2

Answer: x = √2 (repeated root)

Question 14: If the roots of x² + x + k = 0 are real and equal, find the value of k.

Solution:

Step 1: For real and equal roots, discriminant = 0

Step 2: Here a = 1, b = 1, c = k

Step 3: Discriminant = 1² - 4(1)(k) = 1 - 4k

Step 4: Setting 1 - 4k = 0

Step 5: k = 1/4

Answer: k = 1/4

Question 15: The product of two consecutive positive integers is 56. Form a quadratic equation and find the integers.

Solution:

Step 1: Let the integers be x and (x + 1)

Step 2: Product = x(x + 1) = 56

Step 3: x² + x = 56

Step 4: x² + x - 56 = 0

Step 5: Factor: (x + 8)(x - 7) = 0

Step 6: x = -8 or x = 7

Step 7: Since integers are positive, x = 7

Answer: Equation: x² + x - 56 = 0; Integers: 7 and 8

Question 16: Solve the equation x² + 6x + 9 = 0 by factorization.

Solution:

Step 1: Notice this is a perfect square trinomial

Step 2: x² + 6x + 9 = (x + 3)² = 0

Step 3: Therefore x + 3 = 0

Step 4: x = -3

Answer: x = -3 (repeated root)

Question 17: Find the value of k for which the equation x² + (k - 2)x + k = 0 has equal roots.

Solution:

Step 1: For equal roots, discriminant = 0

Step 2: Here a = 1, b = (k - 2), c = k

Step 3: Discriminant = (k - 2)² - 4(1)(k) = 0

Step 4: k² - 4k + 4 - 4k = 0

Step 5: k² - 8k + 4 = 0

Step 6: Using quadratic formula: k = (8 ± √(64 - 16))/2 = (8 ± √48)/2 = (8 ± 4√3)/2 = 4 ± 2√3

Answer: k = 4 + 2√3 or k = 4 - 2√3

Question 18: If the roots of the quadratic equation x² + px + q = 0 are twice the roots of x² + ax + b = 0, find the relation between p, q, a, b.

Solution:

Step 1: Let roots of x² + ax + b = 0 be α, β

Step 2: Then α + β = -a and αβ = b

Step 3: Roots of x² + px + q = 0 are 2α, 2β

Step 4: Sum: 2α + 2β = 2(α + β) = 2(-a) = -2a

Step 5: So -p = -2a, therefore p = 2a

Step 6: Product: (2α)(2β) = 4αβ = 4b

Step 7: So q = 4b

Answer: p = 2a and q = 4b

Question 19: Solve for x: x² + 2x + 2 = 0 and express the roots in the form a + ib.

Solution:

Step 1: Use quadratic formula with a = 1, b = 2, c = 2

Step 2: Discriminant = 4 - 4(1)(2) = 4 - 8 = -4

Step 3: x = (-2 ± √(-4))/(2×1) = (-2 ± 2i)/2 = -1 ± i

Answer: x = -1 + i or x = -1 - i

Question 20: If α and β are the roots of x² + 3x - 10 = 0, form a quadratic equation whose roots are α + 1 and β + 1.

Solution:

Step 1: From x² + 3x - 10 = 0: α + β = -3, αβ = -10

Step 2: New roots are (α + 1) and (β + 1)

Step 3: Sum of new roots = (α + 1) + (β + 1) = α + β + 2 = -3 + 2 = -1

Step 4: Product of new roots = (α + 1)(β + 1) = αβ + α + β + 1 = -10 + (-3) + 1 = -12

Step 5: Required equation: x² - (sum)x + product = 0

Step 6: x² - (-1)x + (-12) = 0

Answer: x² + x - 12 = 0