In each case, find whether the trinomial is a perfect square or not :

$(i) x^{2} + 14 x + 49$ $(i i) a^{2} - 10 a + 25$

$(i i i) 4 x^{2} + 4 x + 1$ $(i v) 9 b^{2} + 12 b + 16$

$(v) 16 x^{2} - 16 x y + y^{2}$ $(v i) x^{2} - 4 x + 16$

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Solution

$(i) x^{2} + 14 x + 49 = (x)^{2} + 2 \times x \times 7 + (7)^{2} = (x + 7)^{2} [∵ a^{2} + 2 a b + b^{2} = (a + b)^{2}] ∴ The given trinomial x^{2} + 14 x + 49 is a perfect square. (i i) a^{2} - 10 a + 25 = (a)^{2} - 2 \times a \times 5 + (5)^{2} = (a - 5)^{2} [∵ a^{2} - 2 a b + b^{2} = (a - b)^{2}] ∴ The given trinomial a^{2} - 10 a + 25 is a perfect square. (i i i) 4 x^{2} + 4 x + 1 = (2 x)^{2} + 2 \times 2 x \times 1 + (1)^{2} = (2 x + 1)^{2} [∵ a^{2} + 2 a b + b^{2} = (a + b)^{2}] ∴ The given trinomial 4 x^{2} + 4 x = 1 is a perfect square. (i v) 9 b^{2} + 12 b + 16 = (3 b)^{2} + 3 b \times 4 + (4)^{2} = x^{2} + x y + y^{2} [Taking 3 b = x, a n d 4 = y] ∴ The given trinomial cannot be expressed as x^{2} + 2 x y + y^{2} . Hence, it is not a perfect square. (v) 16 x^{2} - 16 x y + y^{2} = (4 x)^{2} - 4 \times 4 x \times y + (y)^{2} = a^{2} - 4 a b + b^{2} [Taking 4 x = a, a n d y = b] ∴ The given trinomial cannot be expressed as a^{2} - 2 a b + b^{2} . ∴ It is not a perfect square. (v i) x^{2} - 4 x + 16 = (x)^{2} - x \times 4 + (4)^{2} = a - a b + b^{2} [Taking x = a, a n d 4 = b] ∴ The given trinomial cannot be expressed as a^{2} - 2 a b + b^{2} . Hence, it is not a perfect square.$

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Similar questions

16 x^{2} - 16 x y + y^{2}

9 b^{2} + 12 b + 16

x^{2} - 4 x + 16

Factorise:

(i) x² + 14x + 49; (ii) 4x² + 4x + 1; (iii) a² − 10a + 25; (iv) 2x² − 24x + 72;

(v) p² − 24p + 144; (vi) x³ − 12x² + 36x.

\sqrt{x^{2} - 4 x + 3} + \sqrt{x^{2} - 9} = \sqrt{4 x^{2} - 14 x + 16}

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