RD Sharma Class 9 Solutions Chapter 5 VSAQS (Updated for 2021-22)

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Question 1: Factorize x⁴ + x² + 25

Solution:

x⁴ + x² + 25

= (x²) ² + 5² + x²

[using a² + b² = (a + b) ² – 2ab ]

= (x² +5) ² −2(x² ) (5) + x²

= (x² +5) ² −10x² + x²

= (x² + 5) ² − 9x²

= (x² + 5) ² − (3x) ²

[using a² – b² = (a + b)(a – b ]

= (x ² + 3x + 5)(x² − 3x + 5)

Question 2: Factorize x² – 1 – 2a – a²

Solution:

x² – 1 – 2a – a²

x² – (1 + 2a + a² )

x² – (a + 1)²

(x – (a + 1)(x + (a + 1)

(x – a – 1)(x + a + 1)

[using a² – b² = (a + b)(a – b) and (a + b)^2 = a^2 + b^2 + 2ab ]

Question 3: If a + b + c =0, then write the value of a³ + b³ + c³.

Solution:

We know, a³ + b³ + c³ – 3abc = (a + b +c ) (a² + b² + c² – ab – bc − ca)

Put a + b + c =0

This implies

a³ + b³ + c³ = 3abc

Question 4: If a² + b² + c² = 20 and a + b + c =0, find ab + bc + ca.

Solution:

We know, (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

0 = 20 + 2(ab + bc + ca)

-10 = ab + bc + ca

Or ab + bc + ca = -10

Question 5: If a + b + c = 9 and ab + bc + ca = 40, find a² + b² + c² .

Solution:

We know, (a+b+c)² = a² + b² + c² + 2(ab + bc + ca)

9² = a² + b² + c² + 2(40)

81 = a² + b² + c² + 80

⇒ a² + b² + c² = 1

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