Question

solve the following equation using the quadratic formula:[2÷(x+1)]=[5x÷(3-x)]

Answer 1

hey mate
here's the solution

Answer 2

Given equation : $\dfrac{2}{x+1}=\dfrac{5x}{3-x}$


$\implies \bigg\{ \dfrac{2}{x + 1} \times \dfrac{3 - x}{3 - x} \bigg \} = \bigg \{ \dfrac{5x}{3 - x} \times \dfrac{x + 1}{x + 1} \bigg \} \\ \\ \\ \implies \dfrac{2(3 - x)}{(x + 1)(3 - x)} = \dfrac{5x(x + 1)}{(3 - x)(x + 1)} \\ \\ \\ \implies \dfrac{2(3 - x)}{(x + 1)(3 - x)} - \dfrac{5x(x + 1)}{(3 - x)(x + 1)} = 0$

$\implies \dfrac{ 2(3 - x) -5x( x + 1) }{ (x + 1)(3 + x) } = 0 \\ \\ \\ \implies 2(3 - x) -5x( x + 1) = 0$


$\implies 6 - 2x - 5x^2 - 5x = 0 \\\\\implies -5x^2- 5x - 2x + 6 = 0 \\\\\implies 5x^2 + 7x - 6 = 0$



On Comparing the formed equation with ax² + bx + c = 0, we get that a = 5 , b = 7 ,c = - 6



Now, by Shri Dharra Acharya's method ( Quadratic formula )

$\longrightarrow x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac}}{2a}$




Substituting the values which are given in the question.


$\implies x = \dfrac{ - 7 \pm \sqrt{(7) {}^{2} - 4( - 6 \times 5)} }{2 \times 5} \\ \\ \\ \implies x = \frac{ - 7 \pm \sqrt{49 + 120} }{10} \\ \\ \implies x = \dfrac{ - 7 + \sqrt{169} }{10} \: \: or \: \: \dfrac{ - 7 - \sqrt{169} }{10}$

x = ( - 7 + 13 ) / 10 or ( - 7 - 13 ) / 10

x = 6 / 10 or - 20 / 10

x = 3 / 5 or - 2




Therefore the value of x satisfying the given equation is 3 / 5 or - 2 .