Question

Find the possible value of m if (x²+3x+5)(x²-3x+5)=m²-n². options.(a): x²-3x. (b): 3x. (c): x²+5. (d): x²+2x+1​

Answer 1

option (c) x² + 5

$( {x}^{2} + 3x + 5)( {x}^{2} - 3x + 5) = {m}^{2} - {n}^{2} \\ (( {x}^{2} + 5) + 3x)(( {x}^{2} + 5) - 3x) = {m}^{2} - {n}^{2} \\ {( {x}^{2} + 5) }^{2} - {(3x)}^{2} = {m}^{2} - {n}^{2} \\ {m}^{2} = ( {x}^{2} + 5) }^{2} \\ m = {x}^{2} + 5$

Answer 2

Given that , ( x² + 3x + 5 ) ( x² - 3x + 5 ) = m² - n² .

Exigency To Find : The Possible value for m ?

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\qquad:\implies \sf \Big\{ x ^2 + 3x + 5 \Big\} \:\:\Big\{ x ^2 - 3x + 5 \Big\} \:=\: m^2 - n^2 \:\\\\ \qquad:\implies \sf \Big\{ ( x^2 + 5 ) \:-  3x  \Big\}\:\: \Big\{( x^2+ 5 ) \: - 3x  \Big\}  \:=\:m^2 - n^2 \:$

As , We know that ,

$\qquad \dag\:\:\bigg\lgroup \sf{ Algebraic \:Indentity \::\:a^2 - b^2 = ( a + b ) ( a - b ) }\bigg\rgroup$

$\qquad:\implies \sf \Big\{ ( x^2  + 5 ) \:-  3x  \Big\}\:\: \Big\{  ( x^2  + 5 ) \: - 3x  \Big\}  \:=\:m^2 - n^2 \:\\\\\qquad:\implies \sf \Big\{ ( x^2  + 5 ) ^2 \:-  (3x)^2  \Big\}\:\:   \:=\:m^2 - n^2 \:$

⠀⠀⠀⠀⠀▪︎⠀⠀By Comparing L.H.S to R.H.S :

$\qquad:\implies \sf \Big\{ ( x^2  + 5 ) ^2 \:- (3x)^2 \Big\}\:\:   \:=\:m^2 - n^2$

We get,

$\qquad \dashrightarrow \sf m^2 \:=\: \Big( x^2  + 5 \: \Big)^2\:\:\\\\ \qquad \dashrightarrow \sf m \:=\:\sqrt{ \Big( x^2  + 5 \: \Big)^2\:}\:\\\\ \qquad \dashrightarrow \sf m \:=\: x^2  + 5 \: \:\:\\\\\qquad \dashrightarrow \pmb{\bf{\purple { m \:=\:x^2  + 5}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀AND ,

$\qquad \dashrightarrow \sf n^2 \:=\: \Big( 3x \: \Big)^2\:\:\\\\ \qquad \dashrightarrow \sf n \:=\:\sqrt{ \Big( 3x \: \Big)^2\:}\:\\\\ \qquad \dashrightarrow \sf n \:=\: 3x \: \:\:\\\\\qquad \dashrightarrow \pmb{\bf{\purple { n \:=\:3x }}}$

$\qquad \therefore \:\underline {\sf Hence, \: The \: Possible \: value \: for \: m \: is \: \pmb{\bf Option \: C \:)\:: x^2  + 5}\:.}$