Question

Find the positive value of the variable for which the given equation is satisfied . pls give this answer fast guys .​

Answer 1

Answer:-

The Required Value Of x is 6.

Explanation:-

Given:-

• An equation $\bf\dfrac{3-x^2}{8+x^2} = \dfrac{-3}{4}.$

To Find:-

• The positive value of variable i.e. x.

So,

Let us Solve the given equation,

$\\ \rm\mapsto \frac{3 - {x}^{2} }{8 + {x}^{2} } = \dfrac{ - 3}{4}.$

$\\ \rm\mapsto4(3 - {x}^{2} ) = - 3(8 + {x}^{2}). \: \\ \bf (cross \: \: multiplying).$

$\\ \rm\mapsto12 - 4 {x}^{2} = - 24 - 3 {x}^{2} . \\ \bf(op en in g \: \: brackets).$

$\\ \rm\mapsto12 + 24 = - 3 {x}^{2} + 4 {x}^{2} . \\ \bf(transporting).$

$\\ \rm\mapsto36 = {x}^{2} .$

$\\ \rm\mapsto {x}^{2} = 36.$

$\\ \rm\mapsto x = \pm\sqrt{36}.$

$\\ \rm\mapsto x = \pm\sqrt{6 \times 6} .$

$\\ \large \red{\mapsto \boxed{ \blue{ \bf x = 6\:\:\:\:or\:\:\:\:x= -6.}}}$

But we need to find out the positive value of x.

$\bf\therefore$ The Required Value Of x is 6.

Verification:-

Let put the value of x in the equation to verify the solution,

$\\ \tt\mapsto \frac{3 - {x}^{2} }{8 + {x}^{2} } = \frac{ - 3}{4} .$

$\\ \tt\mapsto \frac{3 - (6) {}^{2} }{8 + (6)^{2} } = \frac{ - 3}{4} .$

$\\ \tt\mapsto \frac{3 - 36}{8 \times 36} = \frac{ - 3}{4} .$

$\\ \tt\mapsto \cancel \frac{ - 33}{44} = \frac{ - 3}{4} .$

$\\ \large \red{\mapsto \boxed{ \blue{ \bf \frac{ - 3}{4} = \frac{ - 3}{4} .}}}$

So LHS = RHS, Hence Verified.