RD Sharma Solutions for Class 11 Mathematics Chapter 14 Quadratic Equations

RD Sharma Solutions for Class 11 Mathematics Chapter 14 “Quadratic Equations” are available here, solved by expert teachers. These solutions strictly follow the latest CBSE Class 11 Maths syllabus and NCERT Book guidelines. On this page, students will find a wide variety of quadratic equations problems with their step-by-step answers. This will help students understand key concepts such as the standard form of a quadratic equation, the quadratic formula, and the nature of roots. It will also help you practice different methods of solving quadratic equations, including factorization, completing the square, and using the quadratic formula. You can revise your full syllabus and easily score higher marks in your exams.

In the RD Sharma Solutions for Class 11 Maths Chapter 14, Students will find detailed, stepwise solutions to all textbook exercises on Quadratic Equations. Special focus is given to important topics like the discriminant, real and complex roots, and the relationship between the roots and coefficients. These fundamental concepts are crucial for solving a wide range of problems in both school-level examinations and competitive entrance tests like JEE Main. The chapter builds upon your previous knowledge of algebra and equips you with the techniques to solve quadratic equations efficiently, analyze the nature of their roots, and apply quadratic equations to real-life word problems.

Download RD Sharma Class 11 Chapter 14 Quadratic Equations Solutions PDF Free

Get comprehensive RD Sharma Class 11 Maths solutions for Quadratic Equations, including step-by-step answers, solved examples, and extra practice questions to help you master quadratic equations and their applications. Learn how to identify the standard form of a quadratic equation, use the quadratic formula, and determine the nature of roots using the discriminant. Download a free PDF to enhance your Class 11 Maths preparation with trusted and exam-oriented RD Sharma Solutions.

RD Sharma Solutions for Quadratic Equations Solved Examples

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

Solution:

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

2x² - 5x + 3 = 0

Step 2: Identify coefficients

a = 2 (coefficient of x²)

b = -5 (coefficient of x)

c = 3 (constant term)

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

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

Step 2: Factor the equation

x² - 7x + 12 = (x - 3)(x - 4) = 0

Step 3: Solve for x

x - 3 = 0 or x - 4 = 0

Therefore, 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

x = [-2 ± √(4 - 4(3)(-1))] / 2(3)

x = [-2 ± √(4 + 12)] / 6

x = [-2 ± √16] / 6

x = [-2 ± 4] / 6

Step 3: Calculate both roots

x = (-2 + 4) / 6 = 2/6 = 1/3

x = (-2 - 4) / 6 = -6/6 = -1

Therefore, 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

Δ = 4² - 4(1)(5) = 16 - 20 = -4

Step 3: Determine nature of roots

Since Δ < 0, the roots are complex (imaginary) 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: Identify coefficients

a = 1, b = -6, c = 8

Step 2: Use Vieta's formulas

Sum of roots: α + β = -b/a = -(-6)/1 = 6

Product of roots: αβ = c/a = 8/1 = 8

Step 3: Verification by solving

x² - 6x + 8 = (x - 2)(x - 4) = 0

So α = 2, β = 4

α + β = 2 + 4 = 6 ✓

αβ = 2 × 4 = 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 side

x² - (3/2)x = -7/2

Step 3: Complete the square

Add (3/4)² = 9/16 to both sides

x² - (3/2)x + 9/16 = -7/2 + 9/16

(x - 3/4)² = -56/16 + 9/16 = -47/16

Step 4: Solve for x

x - 3/4 = ±√(-47/16) = ±i√47/4

x = 3/4 ± i√47/4

The roots are complex: x = (3 ± i√47)/4

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

Solution:

Step 1: Use the relationship (x - α)(x - β) = 0

Where α = 2 and β = -3

Step 2: Substitute and expand

(x - 2)(x - (-3)) = 0

(x - 2)(x + 3) = 0

x² + 3x - 2x - 6 = 0

x² + x - 6 = 0

Therefore, the required quadratic equation is 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

3² + p(3) + 12 = 0

9 + 3p + 12 = 0

21 + 3p = 0

3p = -21

p = -7

Step 2: Find the other root using product of roots

For x² - 7x + 12 = 0, product of roots = 12

If one root is 3, then 3 × (other root) = 12

Other root = 12/3 = 4

Therefore, p = -7 and the other root is 4

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

Solution:

Step 1: Find roots of x² - 5x + 6 = 0

x² - 5x + 6 = (x - 2)(x - 3) = 0

Roots are 2 and 3

Step 2: Find reciprocals

Reciprocals are 1/2 and 1/3

Step 3: Form equation with roots 1/2 and 1/3

Sum of reciprocal roots = 1/2 + 1/3 = 5/6

Product of reciprocal roots = (1/2)(1/3) = 1/6

Step 4: Use x² - (sum)x + product = 0

x² - (5/6)x + 1/6 = 0

Multiply by 6: 6x² - 5x + 1 = 0

Therefore, the required equation is 6x² - 5x + 1 = 0

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

Solution:

Step 1: Check if it's a perfect square

4x² + 4x + 1 = (2x)² + 2(2x)(1) + 1² = (2x + 1)²

Step 2: Solve (2x + 1)² = 0

2x + 1 = 0

2x = -1

x = -1/2

Step 3: Since it's a perfect square, both roots are equal

Therefore, x = -1/2 (repeated root)

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

Solution:

Step 1: Condition for equal roots

For a quadratic equation to have equal roots, its discriminant must be zero

Step 2: Calculate discriminant

Discriminant Δ = b² - 4ac

Step 3: Set discriminant equal to zero

For equal roots: Δ = 0

b² - 4ac = 0

b² = 4ac

Hence proved that for equal roots, b² = 4ac

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

Solution:

Step 1: Identify coefficients

a = 5, b = -2, c = 1

Step 2: Apply Vieta's formulas

Sum of roots = -b/a = -(-2)/5 = 2/5

Product of roots = c/a = 1/5

Therefore, sum of roots = 2/5 and product of roots = 1/5

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

Solution:

Step 1: Identify coefficients

a = 1, b = -2√2, c = 2

Step 2: Calculate discriminant

Δ = b² - 4ac = (-2√2)² - 4(1)(2) = 8 - 8 = 0

Step 3: Since Δ = 0, roots are equal

x = -b/2a = -(-2√2)/2(1) = 2√2/2 = √2

Therefore, 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: Identify coefficients

a = 1, b = 1, c = k

Step 2: For real and equal roots, discriminant = 0

Δ = b² - 4ac = 0

1² - 4(1)(k) = 0

1 - 4k = 0

4k = 1

k = 1/4

Therefore, 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 consecutive integers be x and (x + 1)

Step 2: Form the equation

x(x + 1) = 56

x² + x = 56

x² + x - 56 = 0

Step 3: Solve by factorization

Find two numbers that multiply to -56 and add to 1

The numbers are 8 and -7

x² + 8x - 7x - 56 = 0

x(x + 8) - 7(x + 8) = 0

(x - 7)(x + 8) = 0

Step 4: Find solutions

x = 7 or x = -8

Since we need positive integers, x = 7

Therefore, the consecutive integers are 7 and 8

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

Solution:

Step 1: Recognize it as a perfect square

x² + 6x + 9 = x² + 2(3x) + 3² = (x + 3)²

Step 2: Solve (x + 3)² = 0

x + 3 = 0

x = -3

Therefore, 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: Identify coefficients

a = 1, b = (k - 2), c = k

Step 2: For equal roots, discriminant = 0

Δ = b² - 4ac = 0

(k - 2)² - 4(1)(k) = 0

k² - 4k + 4 - 4k = 0

k² - 8k + 4 = 0

Step 3: Solve for k using quadratic formula

k = [8 ± √(64 - 16)]/2 = [8 ± √48]/2 = [8 ± 4√3]/2 = 4 ± 2√3

Therefore, 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 and β

Then: α + β = -a and αβ = b

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

Sum: 2α + 2β = 2(α + β) = 2(-a) = -2a

Product: (2α)(2β) = 4αβ = 4b

Step 3: Using Vieta's formulas for x² + px + q = 0

Sum of roots = -p, so -p = -2a, therefore p = 2a

Product of roots = q, so q = 4b

Therefore, the relations are: 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: Identify coefficients

a = 1, b = 2, c = 2

Step 2: Calculate discriminant

Δ = b² - 4ac = 4 - 4(1)(2) = 4 - 8 = -4

Step 3: Since Δ < 0, roots are complex

x = [-b ± √Δ]/2a = [-2 ± √(-4)]/2

x = [-2 ± 2i]/2 = -1 ± i

Step 4: Express in the form a + ib

x = -1 + i or x = -1 - i

Therefore, the roots are -1 + i and -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: Find sum and product of original roots

For x² + 3x - 10 = 0:

α + β = -3 and αβ = -10

Step 2: Find sum and product of new roots

New roots are (α + 1) and (β + 1)

Sum: (α + 1) + (β + 1) = α + β + 2 = -3 + 2 = -1

Product: (α + 1)(β + 1) = αβ + α + β + 1 = -10 + (-3) + 1 = -12

Step 3: Form the quadratic equation

Using x² - (sum)x + product = 0

x² - (-1)x + (-12) = 0

x² + x - 12 = 0

Therefore, the required quadratic equation is x² + x - 12 = 0