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The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula itself is:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
The formula helps find the roots (solutions) of quadratic equations. Here is how it works:

• Identify the coefficients: \(a\), \(b\), and \(c\). These come from the quadratic equation.
• Substitute these coefficients into the quadratic formula.
• Calculate the discriminant which is under the square root: \( \Delta = b^2 - 4ac \).
• Evaluate the expression carefully to find the values of \(x\). This might provide two, one, or no real solutions, depending on the discriminant.

By applying the quadratic formula to the equation \( u^2 + 3u - 4 = 0 \), we determined \( u \)'s possible values. First, we noticed that \( a = 1 \), \( b = 3 \), and \(c = -4 \). Next, we plugged these into the formula to solve for \( u \). This resulted in two possible values for \( u \): \( u = 1 \) and \( u = -4 \). The quadratic formula is especially handy because it always works, no matter how complicated the quadratic equation looks. When using it, it's crucial to be cautious with your algebraic steps to ensure accuracy.

The substitution method is a way to simplify and solve equations by replacing one variable or expression with another.
Here is how it generally works:

• Identify a part of the equation that can be substituted with a new variable (often done to simplify a complex equation).
• Replace that part with the new variable.
• Solve the simpler equation for the new variable.
• Substitute back the original expression and solve for the remaining variable.

In our exercise, we used substitution to simplify a complex equation. We let \( u = y + \frac{2}{y} \) since this expression appeared multiple times. This helped transform the equation \( y^2 + \frac{4}{y^2} + 3y + \frac{6}{y} = 0 \) into a simpler quadratic form, \( u^2 + 3u - 4 = 0 \). After solving for \( u \) using the quadratic formula, we translated \( u \) back into the expression \( y + \frac{2}{y} \) to find \(y\). Substitution effectively breaks down complex problems into simpler, more manageable parts. By using it, we turned a seemingly unapproachable equation into a solvable quadratic.

The discriminant is a part of the quadratic formula which helps determine the nature of the roots of a quadratic equation.
It is given by the expression \( \Delta = b^2 - 4ac \). The discriminant tells a lot about the quadratic equation's solutions:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (a repeated root).
• If \( \Delta < 0 \), there are no real solutions (the solutions are complex numbers).

In our equation-solving process, we calculated the discriminants for \( y + \frac{2}{y} = 1 \) and \( y + \frac{2}{y} = -4 \). For \( y + \frac{2}{y} = 1 \), the discriminant was negative, which meant no real solutions for that equation. For \( y + \frac{2}{y} = -4 \), the discriminant was positive, providing two real solutions: \( y = -1 + \sqrt{2} \) and \( y = -1 - \sqrt{2} \). Understanding the discriminant is crucial since it quickly informs us about the type of solutions to expect, simplifying our solving strategy early on.