RD Sharma Class 9 Solutions Chapter 4 Exercise 4.2 (Updated for 2021-22)

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RD Sharma Class 9 Solutions Chapter 4 Exercise 4.2

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Question 1.
Write the following in the expanded form:

Solution:

Question 2.
If a + b + c = 0 and a² + b² + c² = 16, find the value of ab + be + ca.
Solution:
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 =-\(\frac { 16 }{ 2 }\) = -8
∴ ab + bc + ca = -8

Question 3.
If a² + b² + c² = 16 and ab + bc + ca = 10, find the value of a + b + c.
Solution:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
= 16 + 2 x 10
= 16 + 20 = 36
= (±6)²
∴ a + b + c = ±6

Question 4.
If a + b + c = 9 and ab + bc + ca = 23, find the value of a² + b² + c².
Solution:
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
⇒ (9)² = a² + b² + c² + 2 x 23
⇒ 81= a² + b² + c² + 46
⇒  a² + b² + c² = 81 – 46 = 35
∴ a² + b² + c² = 35

Question 5.
Find the value of 4x² + y² + 25z² + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.
Solution:
x = 4, y – 3, z = 2
⇒ 4x² + y² + 25z² + 4xy – 10yz – 20zx
= (2x)² + (y)² + (5z)² + 2 x2 x x y-2 x y x 5z – 2 x 5z x 2x
= (2x + y- 5z)²
= (2 x 4 + 3- 5 x 2)²
= (8 + 3- 10)²
= (11 – 10)²
= (1)² = 1

Question 6.
Simplify:
(i)  (a + b + c)² + (a – b + c)²
(ii) (a + b + c)² –  (a – b + c)²
(iii) (a + b + c)² +   (a – b + c)² + (a + b – c)²
(iv) (2x + p – c)² – (2x – p + c)²
(v) (x² + y² – z²)² – (x² – y² + z²)²
Solution:

Question 7.
Simplify each of the following expressions:

Solution:

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