Show that :

$(i) {(3 x + 7)}^{2} - 84 x = {(3 x - 7)}^{2}$

$(i i) {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 48 p q^{2}$

$(i i i) (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$ [3 MARKS]

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Solution

Application: 1 Mark each

$(i) (3 x + 7)^{2} - 84 x \Rightarrow L . H . S$

$\Rightarrow (3 x)^{2} + 7^{2} + 2 (3 x) (7) - 84 x$

$\Rightarrow 9 x^{2} + 49 + 42 x - 84 x$

$\Rightarrow 9 x^{2} + 49 - 42 x = L . H . S$

$R . H . S \Rightarrow (3 x - 7)^{2} \Rightarrow [(3 x)^{2} + (7)^{2} - 2 (3 x) 7]$

$\Rightarrow 9 x^{2} + 49 - 42 x = R . H . S$

$L . H . S = R . H . S$

$(i i) (4 p q + 3 q)^{2} - (4 p q - 3 q)^{2} = 48 p q^{2}$

$L . H . S \Rightarrow [(4 p q + 3 q) + (4 p q - 3 q)] [((4 p q + 3 q) - (4 p q - 3 q))]$

$\Rightarrow [4 p q + 3 q + 4 p q - 3 q] [4 p q + 3 q - 4 p q + 3 q]$

$\Rightarrow (8 p q) (6 q)$

$\Rightarrow 48 p q^{2}$

$\Rightarrow R . H . S$

$(i i i) (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$

$L . H . S \Rightarrow [(a)^{2} - (b)^{2}] + [(b)^{2} - (c)^{2}] + [(c)^{2} - a^{2}]$

$\Rightarrow a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}$

$\Rightarrow 0 = R . H . S$

Suggest Corrections

Similar questions

(i) {(3 x + 7)}^{2} - 84 x = {(3 x - 7)}^{2}

(i i) {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 48 p q^{2}

(i i i) (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0

{(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16}

Show that

(i) (3x + 7)² − 84x = (3x − 7)²(ii) (9p − 5q)² + 180pq = (9p + 5q)²

(iii)

(iv) (4pq + 3q)²− (4pq − 3q)² = 48pq²

(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

(3 x + 7)^{2} - 84 x = (3 x - 7)^{2}

(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

(4 p q + 3 q)^{2} - (4 p q - 3 q)^{2} = 48 p q^{2}

(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0

Show that:

$(4 p q + 3 q)^{2} - (4 p q - 3 q)^{2} = 48 p q^{2}$

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