Iii Show that- 4-3m-3-4n 2-2mn-16-9m2-9-16n2

L.H.S=( 4/3m 3/4n )^2+2mn= ( 4/3m )^2 2×4/3m×3/4n+ ( 3/4n )^2+2mn [ Using identity (a b)^2=a^2 2ab+b^2 ] =16/9m^2 2mn+9/16n^2+2mn =16/9m^2+9/16n^2 ...

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Standard VIII

Mathematics

Algebraic Identities

iii Show that...

Question

Question 5(iii)
Show that:
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

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Solution

L.H.S= ${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = {(\frac{4}{3} m)}^{2} - 2 \times \frac{4}{3} m \times \frac{3}{4} n + {(\frac{3}{4} n)}^{2} + 2 m n$
$[U s i n g i d e n t i t y (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= \frac{16}{9} m^{2} - 2 m n + \frac{9}{16} n^{2} + 2 m n$
$= \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

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Similar questions

Q. Show that:

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

Question 5(iii)
Show that:
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

Q. Prove that

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

Q. Solve

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

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

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Algebraic Identities

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Question 5(iii) Show that: (43m−34n)2+2mn=169m2+916n2

L.H.S=(43m−34n)2+2mn=(43m)2−2×43m×34n+(34n)2+2mn [Using identity (a−b)2=a2−2ab+b2] =169m2−2mn+916n2+2mn =169m2+916n2

Question 5(iii)
Show that:
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$