Question

$\huge\bf\red{Question}$

show that :-

(4pq + 3q)² - (4pq - 3q)² = 48pq²

​

Answer 1

Step-by-step explanation:

4pq + 3q)² – (4pq – 3q)² = 48pq²

As we know,

(a + b)² = a² + b² + 2ab

and

(a – b)² = a² + b² – 2ab

So, we have to apply the above identities in the given equation.

(4pq + 3q)² – (4pq – 3q)² = L.H.S.

= (16p²q² + 9q² + 24pq²) – (16p²q² + 9q² – 24pq²)

= 16p²q² + 9q² + 24pq² – 16p²q² – 9q² + 24pq²

= 24pq² + 24pq²

= 48pq²

= R.H.S.

Answer 2

$\huge\rm\underline\blue {Solution}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

$\large\bf\underline\red{To ~show~ that~ :-}$

$\rm\orange{{(4pq~ + ~3q)}^{2}~- ~{(4pq - 3q)}^{2}~=~ {48pq}^{2}}$

$\rm{ =>~L.H.S ~ = ~{(4pq~ + ~3q)}^{2} ~-~ {(4pq - 3q)}^{2}}$

$\rm{=>~{(4pq)}^{2} ~+ ~2(4pq)(3q)~ +~{ (3q)}^{2}~ - ~[{(4pq)}^{2}~ -~ 2(4pq)(3q) ~+~{ (3q)}^{2}}$

$\rm{=>~{16p}^{2}{q}^{2}~ + ~{24pq}^{2}~ + ~{9q}^{2}~- ~[{16p}^{2}{q}^{2}~-~ {24pq}^{2}~+ ~{9q}^{2}]}$

$\rm{=>~{16p}^{2}{q}^{2}~+~{ 24pq}^{2}~ +~{ 9q}^{2}~ - ~{16p}^{2}{q}^{2}~ +~ {24pq}^{2}~ - ~{9q}^{2}}$

$\bold{=>~{48pq}^{2}~ = ~R.H.S}$

▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃▃

ɧσ℘ε ıt ɧεɭ℘ร

шıtɧ гεɢศгɖร

❤ ศг + ŋศv ❤