Q. 95.0( 3 Votes )

# If 4x^{2} + y^{2} = 40 and xy = 6, find 2x + y.

Answer :

4x^{2} + y^{2} = 40 …Equation (i)

xy = 6

⇒ 4xy = 24 …Equation (ii)

Adding Equation (i) and (ii)

4x^{2} + y^{2} + 2xy = 64

⇒ (2x + y)^{2} = 8^{2}

⇒ 2x + y = ±8

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Find the product:

(2x – 3y)(2x + 3y)(4x^{2} + 9y^{2})

If a, b are rational numbers such that a^{2} + b^{2} + c^{2} – ab – bc – ca = 0, prove that a = b = c.

Use the identity (a + b)(a – b) = a^{2} – b^{2} to find the products:

(i) (x – 6) (x + 6)

(ii) (3x + 5)(3x + 5)

(iii) (2a + 4b)(2a – 4b)

Karnataka Board - Mathematics Part I

Express the following as difference of two squares:

(i) (x + 2z)(2x + z);

(ii) 4(x + 2y)(2x + y);

(iii) (x + 98)(x + 102);

(iv) 505 × 495.

Karnataka Board - Mathematics Part IFind the product:

(2a + 3)(2a – 3)(4a^{2} + 9)

Simplify:

(i) (x + y)^{2} + (x – y)^{2};

(ii) (x + y)^{2} × (x – y)^{2}.

Find the product:

(p + 2) (p – 2)(p^{2} + 4)

Find the product:

(x – 3)(x + 3)(x^{2} + 9)

Evaluate these using identity:

(i) 55 × 45 (ii) 33 × 27

(iii) 8.5 × 9.5 (iv) 102 × 98

Karnataka Board - Mathematics Part IFind the product:

Karnataka Board - Mathematics Part I