## RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2

Question 1.

Write the following in the expanded form:

Solution:

Question 2.

If a + b + c = 0 and a^{2} + b^{2} + c^{2} = 16, find the value of ab + be + ca.

Solution:

a + b+ c = 0

Squaring both sides,

(a + b + c)^{2} = 0

⇒ a^{2} + b^{2} + c^{2} + 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 a^{2} + b^{2} + c^{2} = 16 and ab + bc + ca = 10, find the value of a + b + c.

Solution:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 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 a^{2} + b^{2} + c^{2}.

Solution:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2(ab + bc + ca)

⇒ (9)^{2} = a^{2} + b^{2} + c^{2} + 2 x 23

⇒ 81= a^{2} + b^{2} + c^{2} + 46

⇒ a^{2} + b^{2} + c^{2} = 81 – 46 = 35

∴ a^{2} + b^{2} + c^{2} = 35

Question 5.

Find the value of 4x^{2} + y^{2} + 25z^{2} + 4xy – 10yz – 20zx when x = 4, y = 3 and z = 2.

Solution:

x = 4, y – 3, z = 2

⇒ 4x^{2} + y^{2} + 25z^{2} + 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) (x^{2} + y^{2} – z^{2})^{2} – (x^{2} – y^{2} + z^{2})^{2 }

Solution:

Question 7.

Simplify each of the following expressions:

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.2 are helpful to complete your math homework.

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