Question

answer this question ​

Answer 1

➛$\frac{3 - {x}^{2} }{8 + {x}^{2} } = \frac{ - 3}{4}$

➛$\frac{3 - {x}^{2} }{8 + {x}^{2} } + \frac{ 3}{4} = 0\\ \large\bf\red \star\green{ for \: positive \: soution}$

➛$\frac{3 - {x}^{2} }{8 + {x}^{2} } + \frac{ 3}{4} ≥ 0$

➛ $\frac{12 - {4x}^{2} + 24 + {3x}^{2} }{(8 + {x}^{2} )(4)} \geq 0$

➛$\frac{36 - {x}^{2} }{32 + {4x}^{2} } \geq 0 \\ if \:taken \: 36 - {x}^{2} \geq 0\:..........equ (1)\\ then \\ x \leq 6 \\\large\bf\green{\star from \: given \: option} \\ \red \star \: x = 6 \\ \bf \purple {answer}$

$\bf\orange{Note}$➠

$\bf{We\:count\:0\:as\:positive\:integer}$

therefore,

$36-x^2=0$

$36-6^2=0$

${36-36=0}$

which is true,

hence.......Prooved.

Answer 2

Question :-

★Positive value of the variable for which the given is satisfied $\sf{\dfrac{3-x^2}{8+x^2}=\dfrac{-3}{4}.}$

○ 6

○7

○None of these

Given :

• $\sf{\dfrac{3-x^2}{8+x^2}=\dfrac{-3}{4}}$

To Find :

• Positive value of the variable.

Solution :-

$:\implies{\sf{\dfrac{3-x^2}{8+x^2}=\dfrac{-3}{4}}} \\ \\ : \implies{\sf{ 4(3-x^2) = - 3(8+x^2)}} \\ \\: \implies{\sf{ 12-4x^2 = - 24 - 3x^2}} \\ \\: \implies{\sf{ 12 + 24= - 3x^2+ 4x^2 }} \\ \\: \implies{\sf{36 = x ^2 }} \\ \\ : \implies{\sf{ \sqrt{36} = x }} \\ \\ : \implies{\sf{6 = x }} \\ \\ :\implies{ \boxed{\tt{x = 6 }}}$

$\underline {\bf{\therefore{Positive \: value \: of \: the\: variable \:is \:6.}}}$

Verification :

We have ,

• Positive value of the variable = 6
• Given Equation :$\sf{\dfrac{3-x^2}{8+x^2}=\dfrac{-3}{4}.}$

Substituting the value of x :

$: \implies{\sf{\dfrac{3-6^2}{8+6^2}=\dfrac{-3}{4}}} \\ \\ : \implies{\sf{ \dfrac{3-36}{8+36}=\dfrac{-3}{4}}} \\ \\ :\implies{\sf{ \dfrac{-33}{44}=\dfrac{-3}{4}}} \\ \\ : \implies{\sf{ \dfrac{-3}{4}=\dfrac{-3}{4}}}\\ \\: \implies{\boxed{\tt{LHS = RHS}}}$

$\underline {\bf{\therefore{Verified.}}}$