Show that:(i) (3x+7)2−84x=(3x−7)2(ii) (9p−5q)2+180pq=(9p+5q)2(iii) (34m−43n)2+2mn=916m2+169n2(iv) (4pq+3q)2−(4pq−3q)2=48pq2(v) (a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a)=0

We know that(a±b)2=a2+b2±2aband,(a−b)(a+b)=a2−b2Now,(i) (3x+7)2 −84x=(3x)2+72+2(3x)(7)−84x=(3x)2+72+42x−84x=(3x)2+72−42x=(3x−7)2(ii) (9p−5q)2+180pq=(9p)2+(5q)2−2(9p)(5q)+180pq=(9p)2+(5q)2−90pq+180pq=(9p)2+(5q)2+90pq=(9p+5q)2(iii)(34m−43n)2+2mn=(34m)2+(43n)2−2(34m)(43n)+2mn=(34m)2+(43n)2−2mn+2mn=916m2+169n2(iv)(4pq+3q)2−(4pq−3q)2=(4pq)2+(3q)2+2(4pq)(3q)−((4pq)2+(3q)2−2(4pq)(3q))=(4pq)2+(3q)2+2(4pq)(3q)−(4pq)2−(3q)2+2(4pq)(3q)=24pq2+24pq2=48pq2(v)(a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a)=a2−b2+b2−c2+c2−a2=0

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Show that: (i) (3x+7)^2-84x=(3x-7)^2 (ii) (9p-5q)^2+180pq =

Question

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

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

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

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

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

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Show that:
(i)

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

(ii)

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

(iii)

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

(iv)

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

(v)

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

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Text solutionVerified

We know that

(a \pm b)^{2} = a^{2} + b^{2} \pm 2 ab

and,

(a - b) (a + b) = a^{2} - b^{2}

Now,

(i)

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

= (3 x)^{2} + 7^{2} + 2 (3 x) (7) - 84 x = (3 x)^{2} + 7^{2} + 42 x - 84 x

= (3 x)^{2} + 7^{2} - 42 x = (3 x - 7)^{2}

(ii)

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

= (9 p)^{2} + (5 q)^{2} - 2 (9 p) (5 q) + 180 pq = (9 p)^{2} + (5 q)^{2} - 90 pq + 180 pq

= (9 p)^{2} + (5 q)^{2} + 90 pq = (9 p + 5 q)^{2}

(iii)

(\frac{4}{3} m - \frac{3}{4} n)^{2} + 2 mn

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

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

= \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

(iv)

(4 pq + 3 q)^{2} - (4 pq - 3 q)^{2}

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

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

= 24 p q^{2} + 24 p q^{2} = 48 p q^{2}

(v)

(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)

= a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}

= 0

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Question Text	Show that: (i) $(3 x + 7)^{2} - 84 x = (3 x - 7)^{2}$ (ii) $(9 p - 5 q)^{2} + 180 pq = (9 p + 5 q)^{2}$ (iii) $(\frac{4}{3} m - \frac{3}{4} n)^{2} + 2 mn = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$ (iv) $(4 pq + 3 q)^{2} - (4 pq - 3 q)^{2} = 48 p q^{2}$ (v) $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$
Answer Type	Text solution:1
Upvotes	150

Show that:(i) (3x+7)2−84x=(3x−7)2(ii) (9p−5q)2+180pq=(9p+5q)2(iii) (34​m−43​n)2+2mn=916​m2+169​n2(iv) (4pq+3q)2−(4pq−3q)2=48pq2(v) (a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a)=0

Text solutionVerified

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