Question

solve for x
$2( \frac{2x + 3}{x - 3} ) - 25( \frac{x - 3}{2x + 3} ) = 5$
find roots of X or zeros of x​

Answer 1

Answer:

$\qquad\qquad\boxed{ \bf{ \: \:x = 1, \: 6 \: \: }}$

Step-by-step explanation:

Given equation is

$\sf \: 2\left(\dfrac{2x + 3}{x - 3}\right) - 25\left( \dfrac{x - 3}{2x + 3}\right)= 5$

Let assume that

$\qquad\sf \: \dfrac{2x + 3}{x - 3} = y$

So, above expression can be rewritten as

$\sf \: 2y - \dfrac{25}{y} = 5$

$\sf \: \dfrac{ {2y}^{2} - 25}{y} = 5$

$\sf \: {2y}^{2} - 25 = 5y$

$\sf \: {2y}^{2} - 5y- 25 = 0$

On splitting the middle terms, we get

$\sf \: {2y}^{2} - 10y + 5y- 25 = 0$

$\sf \: 2y(y - 5) + 5(y - 5) = 0$

$\sf \: (y - 5) \: (2y + 5) = 0$

$\sf\implies \: y = 5 \: \: or \: \: y = - \dfrac{5}{2}$

Now, Consider

$\sf \: y = 5$

$\sf \: \dfrac{2x + 3}{x - 3} = 5$

$\sf \: 2x + 3 = 5x - 15$

$\sf \: 2x - 5x = - 15 - 3$

$\sf \: - 3x = - 18$

$\bf\implies \: x = 6$

Now, Consider

$\sf \: y = - \dfrac{5}{2}$

$\sf \: \dfrac{2x + 3}{x - 3} = - \dfrac{5}{2}$

$\sf \: 2(2x + 3) = - 5(x - 3)$

$\sf \:4x +6 = - 5x + 15$

$\sf \:4x + 5x =15 - 6$

$\sf \:9x =9$

$\bf\implies \: x \: = \: 1$

$\rule{190pt}{2pt}$

Additional Information

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac