Show that-i 3 x-72-84 x-3 x-72ii 9 a-5 b2-180 a b-9 a-5 b2iii 4 m 3-3 n 42-2 mn -16 m 29-9 n 216iv 4 p q-3 q2-4 p q-3 q2-48 p q2v a-ba-b-b-cb-c-c-ac-a-0

(i) #160;LHS=3x+72 84x=3x+72 4 #215;3x #215;7=3x 72 #160; #160; #160; #160; #160; #160; #160; #160; #160; #8757; #160;a+b2 4ab=a b2 ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

Show that:i 3...

Question

Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii) ${(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16}$
(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

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Solution

$(i) LHS = {(3 x + 7)}^{2} - 84 x = {(3 x + 7)}^{2} - 4 \times 3 x \times 7 = {(3 x - 7)}^{2} [∵ {(a + b)}^{2} - 4 a b = {(a - b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(ii) LHS = {(9 a - 5 b)}^{2} + 180 a b = {(9 a - 5 b)}^{2} + 4 \times 9 a \times 5 b = {(9 a + 5 b)}^{2} [∵ {(a - b)}^{2} + 4 a b = {(a + b)}^{2}] = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iii) LHS = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = {(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 \times \frac{4 m}{3} \times \frac{3 n}{4} = {(\frac{4 m}{3})}^{2} + {(\frac{3 n}{4})}^{2} [∵ {(a - b)}^{2} + 2 a b = a^{2} + b^{2}] = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(iv) LHS = {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 4 (4 p q) (3 q) [∵ {(a + b)}^{2} - {(a + b)}^{2} = 4 a b] = 48 p q^{2} = RHS Because LHS is equal to RHS, the given equation is verified.$

$(v) LHS = (a - b) (a + b) + (b - c) (b + c) + (c + a) (c - a) = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} [∵ (a + b) (a - b) = a^{2} - b^{2}] = a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2} = 0 = RHS Because LHS is equal to RHS, the given equation is verified.$

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Similar questions

Q. 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

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

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]

Q. Show that:
(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

Show that:

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

Join BYJU'S Learning Program

Show that: (i) (3x + 7)2 − 84x = (3x − 7)2 (ii) (9a − 5b)2 + 180ab = (9a + 5b)2 (iii) 4m3-3n42+2mn=16m29+9n216 (iv) (4pq + 3q)2 − (4pq − 3q)2 = 48pq2 (v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii) ${(\frac{4 m}{3} - \frac{3 n}{4})}^{2} + 2 m n = \frac{16 m^{2}}{9} + \frac{9 n^{2}}{16}$
(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0