RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.1: Chapter 8 of RD Sharma's Class 10 Maths focuses on Quadratic Equations, a fundamental topic in algebra. Exercise 8.1 introduces the basic concepts of quadratic equations, including their standard form.
It emphasizes identifying quadratic equations and distinguishing them from other algebraic expressions. Students practice converting equations into standard form and verifying whether a given equation is quadratic. This exercise lays the groundwork for solving quadratic equations through various methods covered in subsequent exercises. It strengthens conceptual understanding and prepares students for advanced problem-solving.RD Sharma Solutions Class 10 Maths Chapter 8 Exercise 8.1 PDF
1. Which of the following are quadratic equations?
(i) x 2 + 6x – 4 = 0
Solution:
Let p(x) = x 2 + 6x – 4, It’s clearly seen that p(x) = x 2 + 6x – 4 is a quadratic polynomial. Thus, the given equation is a quadratic equation.(ii) √3x 2 – 2x + 1/2 = 0
Solution:
Let p(x) = √3x 2 – 2x + 1/2, It’s clearly seen that p(x) = √3x 2 – 2x + 1/2 having real coefficients is a quadratic polynomial. Thus, the given equation is a quadratic equation.(iii) x 2 + 1/x 2 = 5
Solution:
Given, x 2 + 1/x 2 = 5 On multiplying by x 2 on both sides, we have, x 4 + 1 = 5x 2 ⇒ x 4 – 5x 2 + 1 = 0 It’s clearly seen that x 4 – 5x 2 + 1 is not a quadratic polynomial as its degree is 4. Thus, the given equation is not a quadratic equation.(iv) x – 3/x = x 2
Solution:
Given, x – 3/x = x 2 On multiplying by x on both sides, we have, x 2 – 3 = x 3 ⇒ x 3 – x 2 + 3 = 0 It’s clearly seen that x 3 – x 2 + 3 is not a quadratic polynomial, as its degree is 3. Thus, the given equation is not a quadratic equation.(v) 2x 2 – √(3x) + 9 = 0
Solution:
It’s clearly seen that 2x 2 – √(3x) + 9 is not a polynomial because it contains a term involving x 1/2 , where 1/2 is not an integer. Thus, the given equation is not a quadratic equation.(vi) x 2 – 2x – √x – 5 = 0
Solution:
It’s clearly seen that x 2 – 2x – √x – 5 is not a polynomial because it contains a term involving x 1/2 , where 1/2 is not an integer. Thus, the given equation is not a quadratic equation.(vii) 3x 2 – 5x + 9 = x 2 – 7x + 3
Solution:
Given, 3x 2 – 5x + 9 = x 2 – 7x + 3 On simplifying the equation, we have 2x 2 + 2x + 6 = 0 ⇒ x 2 + x + 3 = 0 (dividing by 2 on both sides) Now, it’s clearly seen that x 2 + x + 3 is a quadratic polynomial. Thus, the given equation is a quadratic equation.(viii) x + 1/x = 1
Solution:
Given, x + 1/x = 1 On multiplying by x on both sides we have, x 2 + 1 = x ⇒ x 2 – x + 1 = 0 It’s clearly seen that x 2 – x + 1 is a quadratic polynomial. Thus, the given equation is a quadratic equation.(ix) x 2 – 3x = 0
Solution:
Let p(x) = x 2 – 3x, It’s clearly seen that p(x) = x 2 – 3x is a quadratic polynomial. Thus, the given equation is a quadratic equation.(x) (x + 1/x) 2 = 3(x + 1/x) + 4
Solution:
Given, (x + 1/x) 2 = 3(x + 1/x) + 4 ⇒ x 2 + 1/x 2 + 2 = 3x + 3/x + 4 ⇒ x 4 + 1 + 2x 2 = 3x 3 + 3x + 4x 2 ⇒ x 4 – 3x 3 – 2x 2 – 3x + 1 = 0 Now, it’s clearly seen that x 4 – 3x 3 – 2x 2 – 3x + 1 is not a quadratic polynomial since its degree is 4. Thus, the given equation is not a quadratic equation.(xi) (2x + 1)(3x + 2) = 6(x – 1)(x – 2)
Solution:
Given, (2x + 1)(3x + 2) = 6(x – 1)(x – 2) ⇒ 6x 2 + 4x + 3x + 2 = 6x 2 -12x – 6x + 12 ⇒ 7x + 2 = -18x + 12 ⇒ 25x – 10 = 0 Now, it’s clearly seen that 25x – 10 is not a quadratic polynomial since its degree is 1. Thus, the given equation is not a quadratic equation.(xii) x + 1/x = x 2 , x ≠ 0
Solution:
Given, x + 1/x = x 2 On multiplying by x on both sides, we have, x 2 + 1 = x 3 ⇒ x 3 – x 2 – 1 = 0 Now, it’s clearly seen that x 3 – x 2 – 1 is not a quadratic polynomial since its degree is 3. Thus, the given equation is not a quadratic equation.(xiii) 16x 2 – 3 = (2x + 5)(5x – 3)
Solution:
Given, 16x 2 – 3 = (2x + 5)(5x – 3) 16x 2 – 3 = 10x 2 – 6x + 25x – 15 ⇒ 6x 2 – 19x + 12 = 0 Now, it’s clearly seen that 6x 2 – 19x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic equation.(xiv) (x + 2) 3 = x 3 – 4
Solution:
Given, (x + 2) 3 = x 3 – 4 On expanding, we get ⇒ x 3 + 6x 2 + 8x + 8 = x 3 – 4 ⇒ 6x 2 + 8x + 12 = 0 Now, it’s clearly seen that 6x 2 + 8x + 12 is a quadratic polynomial. Thus, the given equation is a quadratic equation.(xv) x(x + 1) + 8 = (x + 2)(x – 2)
Solution:
Given,x(x + 1) + 8 = (x + 2)(x – 2)
x 2 + x + 8 = x 2 – 4 ⇒ x + 12 = 0 Now, it’s clearly seen that x + 12 is a not quadratic polynomial since its degree is 1. Thus, the given equation is not a quadratic equation.2. In each of the following, determine whether the given values are solutions of the given equation or not:
(i) x 2 – 3x + 2 = 0 , x = 2 , x = – 1
Solution:
Here we have, LHS = x 2 – 3x + 2 Substituting x = 2 in LHS, we get (2) 2 – 3(2) + 2 ⇒ 4 – 6 + 2 = 0 = RHS ⇒ LHS = RHS Thus, x = 2 is a solution of the given equation. Similarly, Substituting x = -1 in LHS, we get (-1) 2 – 3(-1) + 2 ⇒ 1 + 3 + 2 = 6 ≠ RHS ⇒ LHS ≠ RHS Thus, x = -1 is not a solution of the given equation.(ii) x 2 + x + 1 = 0, x = 0, x = 1
Solution:
Here we have, LHS = x 2 + x + 1 Substituting x = 0 in LHS, we get (0) 2 + 0 + 1 ⇒ 1 ≠ RHS ⇒ LHS ≠ RHS Thus, x = 0 is not a solution of the given equation. Similarly, Substituting x = 1 in LHS, we get (1) 2 + 1 + 1 ⇒ 3 ≠ RHS ⇒ LHS ≠ RHS Thus, x = 1 is not a solution of the given equation.(iii) x 2 − 3√3x + 6 = 0 , x = √3 and x = −2√3
Solution:
Here we have, LHS = x 2 − 3√3x + 6 Substituting x = √3 in LHS, we get (√3) 2 − 3√3(√3) + 6 ⇒ 3 – 9 + 6 = 0 = RHS ⇒ LHS = RHS Thus, x = √3 is a solution of the given equation. Similarly, Substituting x = −2√3 in LHS, we get (-2√3) 2 − 3√3(-2√3) + 6 ⇒ 12 + 18 + 6 = 36 ≠ RHS ⇒ LHS ≠ RHS Thus, x = −2√3 is not a solution of the given equation.(iv) x + 1/x = 13/6, x = 5/6, x = 4/3
Solution:
Here we have, LHS = x +1/ x Substituting x = 5/6 in LHS, we get (5/6) + 1/(5/6) = 5/6 + 6/5 ⇒ 61/30 ≠ RHS ⇒ LHS ≠ RHS Thus, x = 5/6 is not a solution of the given equation. Similarly, Substituting x = 4/3 in LHS, we get (4/3) + 1/(4/3) = 4/3 + 3/4 ⇒ 25/12 ≠ RHS ⇒ LHS ≠ RHS Thus, x = 4/3 is not a solution of the given equation.(v) 2x 2 – x + 9 = x 2 + 4x + 3, x = 2, x = 3
Solution:
Here we have, 2x 2 – x + 9 = x 2 + 4x + 3 ⇒ x 2 – 5x + 6 = 0 LHS = x 2 – 5x + 6 Substituting x = 2 in LHS, we get (2) 2 – 5(2) + 6 ⇒ 4 – 10 + 6 = 0 = RHS ⇒ LHS = RHS Thus, x = 2 is a solution of the given equation. Similarly, Substituting x = 3 in LHS, we get (3) 2 – 5(3) + 6 ⇒ 9 – 15 + 6 = 0 = RHS ⇒ LHS = RHS Thus, x = 3 is a solution of the given equation.(vi) x 2 – √2x – 4 = 0, x = -√2, x = -2√2
Solution:
Here we have, LHS = x 2 – √2x – 4 Substituting x = -√2 in LHS, we get (-√2) 2 − √2(-√2) – 4 ⇒ 4 + 2 – 4 = 2 ≠ RHS ⇒ LHS ≠ RHS Thus, x = -√2 is a solution of the given equation. Similarly, Substituting x = −2√2 in LHS, we get (-2√2) 2 − √2(-2√2) – 4 ⇒ 8 + 4 – 4 = 8 ≠ RHS ⇒ LHS ≠ RHS Thus, x = −2√2 is not a solution of the given equation.(vii) a 2 x 2 – 3abx + 2b 2 = 0, x = a/b, x = b/a
Solution:
We have, LHS = a 2 x 2 – 3abx + 2b 2 and RHS = 01. Strengthens Conceptual Understanding
2. Improves Problem-Solving Skills
3. Prepares for Board Exams
4. Builds Strong Foundation for Advanced Topics
5. Enhances Accuracy and Speed
6. Boosts Confidence