91Ó°ÊÓ

Cross multiplication is a powerful technique used to solve equations involving fractions. When you have an equation like \( \frac{a}{b} = \frac{c}{d} \), you can remove the fractions by multiplying both sides in the following way: \( a \cdot d = b \cdot c \). This step eliminates the fractions, making the equation easier to solve. In our given exercise, we have the equation \( \frac{5x}{14x + 3} = \frac{1}{x} \). To cross multiply, we perform:
\( 5x \cdot x = (14x + 3) \cdot 1 \). This results in: \( 5x^2 = 14x + 3 \).
Now, the equation is free from fractions and can be rearranged to the standard quadratic form.

Once we have our quadratic equation \( 5x^2 - 14x - 3 = 0 \), the next step is to solve for \( x \). One effective way to solve a quadratic equation is by using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
This formula provides a way to find the roots of any quadratic equation in the form \( ax^2 + bx + c = 0 \). For our specific case, the values of \( a \), \( b \), and \( c \) are 5, -14, and -3 respectively.
Substituting these into the formula gives us: \( x = \frac{-(-14) \pm \sqrt{(-14)^2 - 4(5)(-3)}}{2(5)} \). Simplifying that, we get: \( x = \frac{14 \pm \sqrt{196 + 60}}{10} \) which further simplifies to: \( x = \frac{14 \pm 16}{10} \).
Hence, we find two solutions: \( x = 3 \) and \( x = -\frac{1}{5} \).

The discriminant is a vital part of the quadratic formula, and it determines the nature of the roots of a quadratic equation. The discriminant is given by: \( \Delta = b^2 - 4ac \).
For our equation, we calculated the discriminant as follows: \( (-14)^2 - 4(5)(-3) = 196 + 60 = 256 \).
The value of the discriminant tells us a lot about the solutions:

• If the discriminant is positive (\( > 0 \)), the quadratic equation has two distinct real roots.
• If the discriminant is zero (\( = 0 \)), the equation has exactly one real root or a repeated root.
• If the discriminant is negative (\( < 0 \)), the equation has no real roots, but two complex roots.

In our problem, the discriminant is 256, which is positive. This means we have two distinct real solutions: \( x = 3 \) and \( x = -\frac{1}{5} \). Always remember that calculating the discriminant helps in verifying the nature of the solutions without fully solving the equation.