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Algebraic Quadratic Identity

Question 4Exp...

Question

Question 4
Expand each of the following, using suitable identities:
(i) $(x + 2 y + 4 z)^{2}$
(ii) $(2 x - y + z)^{2}$
(iii) $(- 2 x + 3 y + 2 z)^{2}$
(iv) $(3 a - 7 b - c)^{2}$
(v) $(- 2 x + 5 y - 3 z)^{2}$
(vi) ${[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}$

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Solution

$(i) (x + 2 y + 4 z)^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a] = x^{2} + (2 y)^{2} + (4 z)^{2} + 2 x \times 2 y + 2 \times 2 y \times 4 z + 2 \times 4 z \times x x^{2} + 4 y^{2} + 16 z^{2} + 4 x y + 16 y z + 8 z x$

$(i i) (2 x - y + z)^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a] = (2 x)^{2} + (- y)^{2} + (z)^{2} + 2 \times (2 x) (- y) + 2 (- y) (z) + 2 \times 2 x z = 4 x^{2} + y^{2} + z^{2} - 4 x y - 2 y z + 4 z x$

$(i i i) (- 2 x + 3 y + 2 z)^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a] = (- 2 x)^{2} + (3 y)^{2} + (2 z)^{2} + 2 (- 2 x) (3 y) + 2 (3 y) (2 z) + 2 (2 z) (- 2 x) = 4 x^{2} + 9 y^{2} + 4 z^{2} - 12 x y + 12 y z - 8 z x$

$(i v) (3 a - 7 b - c)^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a] = (3 a)^{2} + (- 7 b)^{2} + (- c)^{2} + 2 (3 a) (- 7 b) + 2 (- 7 b) (- c) + 2 (- c) (3 a) = 9 a^{2} + 49 b^{2} + c^{2} - 42 a b + 14 b c - 6 a c$

$(v) (- 2 x + 5 y - 3 z)^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a] = (- 2 x)^{2} + (5 y)^{2} + (- 3 z)^{2} + 2 (- 2 x) (5 y) + 2 (5 y) (- 3 z) + 2 (- 3 z) (- 2 x) = 4 x^{2} + 25 y^{2} + 9 z^{2} - 20 x y - 30 y z + 12 z x$

$(v i) {[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2} [U s i n g t h e i d e n t i t y (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a]$

$\Rightarrow {(\frac{1}{4} a)}^{2} + {(\frac{- 1}{2} b)}^{2} + (1)^{2} + 2 (\frac{1}{4} a) (\frac{- 1}{2} b) + 2 (\frac{- 1}{2} b) \times 1 + 2 (1) \times \frac{1}{4} a = \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 - \frac{1}{4} a b - b + \frac{1}{2} a = \frac{a^{2}}{16} + \frac{b^{2}}{4} + 1 - \frac{1}{4} a b - b + \frac{1}{2} a = \frac{a^{2}}{16} + \frac{b^{2}}{4} + 1 - \frac{a b}{4} - b + \frac{a}{2}$

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Expand the following, using suitable identities:
$1) {(- 2 x + 3 y + 2 z)}^{2}$
$2) {(- 2 x + 5 y - 3 z)}^{2}$ . [4 MARKS]

Find the square of :

(i) $x + 3 y$

(ii) $2 x - 5 y$

(iii) $a + \frac{1}{5 a}$

(iv) $2 a - \frac{1}{a}$

(v) $x - 2 y + 1$

(vi) $3 a - 2 b - 5 c$

(vii) $2 x + \frac{1}{x} + 1$

(viii) $5 - x + \frac{2}{x}$

(ix) $2 x - 3 y + z$

(x) $x + \frac{1}{x} - 1$

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