RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.1

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RD Sharma Solutions for Class 9 Chapter 4 Algebraic Identities Ex 4.1 Download PDF

Question 1.
Evaluate each of the following using identities:
(i) (2x –1x)²
(ii)  (2x + y) (2x – y)
(iii) (a²b – b²a)²
(iv) (a – 0.1) (a + 0.1)
(v) (1.5.x² – 0.3y²) (1.5x² + 0.3y²)
Solution:

Question 2.
Evaluate each of the following using identities:
(i) (399)²
(ii) (0.98)²
(iii) 991 x 1009
(iv) 117 x 83
Solution:

Question 3.
Simplify each of the following:

Solution:

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
If 9x² + 25y² = 181 and xy = -6, find the value of 3x + 5y.
Solution:
9x² + 25y² = 181, and xy = -6
(3x + 5y)² = (3x)² + (5y)² + 2 x 3x + 5y
⇒ 9X² + 25y² + 30xy
= 181 + 30 x (-6)
= 181 – 180 = 1
= (±1 )²
∴ 3x + 5y = ±1

Question 8.
If 2x + 3y = 8 and xy = 2, find the value of 4X² + 9y².
Solution:
2x + 3y = 8 and xy = 2
Now, (2x + 3y)² = (2x)² + (3y)² + 2 x 2x x 3y
⇒  (8)² = 4x² + 9y² + 12xy
⇒ 64 = 4X² + 9y² + 12 x 2
⇒ 64 = 4x² + 9y² + 24
⇒ 4x² + 9y² = 64 – 24 = 40
∴ 4x² + 9y² = 40

Question 9.
If 3x -7y = 10 and xy = -1, find the value of 9x² + 49y²
Solution:
3x – 7y = 10, xy = -1
3x -7y= 10
Squaring both sides,
(3x – 7y)² = (10)²
⇒ (3x)² + (7y)² – 2 x 3x x 7y = 100
⇒  9X² + 49y² – 42xy = 100
⇒  9x² + 49y² – 42(-l) = 100
⇒ 9x² + 49y² + 42 = 100
∴ 9x² + 49y² = 100 – 42 = 58

Question 10.
Simplify each of the following products:

Solution:

Question 11.

Solution:

Question 12.

Solution:

Question 13.
Simplify each of the following products:

Solution:

Question 14.
Prove that a² + b² + c² – ab – bc – ca is always non-negative for all values of a, b and c.
Solution:

∵  The given expression is sum of these squares
∴ Its value is always positive Hence the given expression is always non-­negative for all values of a, b and c

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