(2x/x-5)^2+(2x/x-5)-24=0

2 min read Jun 16, 2024

Solving the Quadratic Equation: (2x/x-5)^2 + (2x/x-5) - 24 = 0

This equation appears complex, but we can simplify it using a substitution technique and then apply the quadratic formula. Here's how:

Let's simplify the equation by substituting y = (2x/x-5). This gives us:

y² + y - 24 = 0

Now we have a standard quadratic equation in terms of 'y'.

We can solve this equation using the quadratic formula:

y = (-b ± √(b² - 4ac)) / 2a

Where a = 1, b = 1, and c = -24.

Substituting the values into the quadratic formula:

y = (-1 ± √(1² - 4 * 1 * -24)) / 2 * 1

y = (-1 ± √(97)) / 2

This gives us two possible solutions for 'y':

y1 = (-1 + √97) / 2

y2 = (-1 - √97) / 2

Now we need to substitute 'y' back with its original value (2x/x-5) and solve for 'x':

For y1:

(2x/x-5) = (-1 + √97) / 2

Solving for 'x' requires cross-multiplication and simplification, leading to:

x1 = (5 * (-1 + √97)) / (2 - √97)

For y2:

(2x/x-5) = (-1 - √97) / 2

Similarly, solving for 'x' gives:

x2 = (5 * (-1 - √97)) / (2 + √97)

Therefore, the solutions for the original equation are:

x1 = (5 * (-1 + √97)) / (2 - √97)

x2 = (5 * (-1 - √97)) / (2 + √97)

These are the two solutions to the given equation.