RD Sharma Chapter 4 Class 9 Maths Exercise 4.2 Solutions

RD Sharma Chapter 4 Class 9 Maths Exercise 4.2 Solutions is based on the Square of a Trinomial, which is the part of the Algebraic Expressions. The answers to the questions are prepared by the proficient of our institute in an easy way, which helps students to use the formulas in the right problems. The tips and tricks are also mentioned in the RD Sharma Chapter 4 Class 9 Maths Exercise 4.2 Solutions PDF to make the solutions easier for the learners.

In Exercise 4.1, we have learned about the standard Algebraic identities. But, in Exercise 4.2, students will learn about some special cases of the square of a trinomial that is mentioned in this article with formulas and examples.

Learn about RD Sharma Class 9 Chapter 4- Algebraic Identities

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Solutions for Class 9 Maths Chapter 4 Algebraic Identities Exercise 4.2

Important Definitions RD Sharma Chapter 4 Class 9 Maths Exercise 4.2 Solutions

Have a look at the formulas of the square of a trinomial mentioned below in the points-

  1. (a+b+c)2= a2+ b2+ c2+ 2ab- 2bc- 2ca
  2. (a-b+c)2= a2+ b2+ c2- 2ab- 2bc+ 2ca
  3. (-a+b+c)2=a2+b2+c2-2ab+2bc-2ca
  4. (a-b-c)2= a2+ b2+ c2- 2ab- 2bc- 2ca

Expand Form of a Square of a Trinomial

The square of the total of three or more expressions can be defined by the formula of the determination of the square of the sum of two expressions.

Now check how to expand a square of a Trinomial-

Let, (b+c) = x

Then,

=(a + b + c)2 = (a + x)2 = a2 + 2ax + x2

= a2 + 2a (b + c) + (b + c)2

= a2 + 2ab + 2ac + (b2 + c2 + 2bc)

= a2 + b2 + c2 + 2ab + 2bc + 2ca

Therefore, (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca has proved.

Examples-

Ques- If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca.

Solution

a + b + c = 0 and a2 + b2 + c2 = 16 (given)

Choose a + b + c = 0

Squaring both sides,

(a + b + c)2 = 0

a2 + b2 + c2 + 2(ab + bc + ca) = 0

16 + 2(ab + bc + c) = 0

2(ab + bc + ca) = -16

ab + bc + ca = -16/2 = -8

or ab + bc + ca = -8

Ques- (a + b + c)2 − (a − b + c) 2

Solution-

(a + b + c)2 − (a − b + c) 2= (a2 + b2 + c2 + 2ab + 2bc + 2ca) − (a2 + (−b) 2 + c2 −2ab − 2bc + 2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca

= 4ab + 4bc

Ques- (p2 + q2 − r2) 2 − (p2 − q2 + r2) 2

Solution-

(p2 + q2 − r2) 2 − (p2 − q2 + r2) 2 = (p2 + q2 + (−r) 2) 2 − (p2 − q2 + r2) 2

= [p4 + q4 + r4 + 2p2q2 – 2q2r 2 – 2p2r2 − [p4 + q4 + r4 − 2p2q2 − 2q2r2 + 2p2r2]

= 4p2q2 – 4r2p2

Frequently Asked Questions (FAQs) of RD Sharma Chapter 4 Class 9 Maths Exercise 4.2 Solutions

Ques 1- How do you find the square of a trinomial?

Ans- Take a binomial and multiply it to itself. You end up with a perfect square trinomial.

Ques 2- How do you complete the square of a trinomial?

Ans- A perfect square trinomial is factored in, so the equation can be solved by taking the square root of both sides.

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