91Ó°ÊÓ

When encountering a quadratic equation, which is any polynomial equation of the second degree, the quadratic formula is a reliable method for finding its solutions. This formula is represented as \( y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \), where \( a \) is the coefficient of the squared term, \( b \) is the coefficient of the linear term, and \( c \) is the constant term.

The beauty of the quadratic formula is that it is universally applicable to all quadratic equations, and it provides both real and complex solutions. To use it effectively, you'll need first to identify the coefficients correctly and then simply substitute them into the formula.

For the exercise given: \( 4s^4 + 11s^2 - 3 = 0 \), when you substitute \( s^2 \) with \( y \) in the first step, it reduces to a quadratic equation. Then you apply the quadratic formula with \( a=4 \) , \( b=11 \) and \( c=-3 \) to find solutions for \( y \).

The substitution method is a technique used in algebra to simplify complex polynomial equations. It involves replacing a variable with another expression to reduce the equation to a more manageable form. This method is especially useful when dealing with higher-degree polynomials that can be transformed into quadratic equations.

In the provided exercise, \( y = s^2 \) is used as a substitution to convert the quartic equation \( 4s^4 + 11s^2 - 3 = 0 \) into a quadratic equation. This creative maneuver simplifies the process of finding the solutions since solving a quadratic equation is a well-understood and straightforward process. After finding the values of \( y \) using the quadratic formula, you then reverse the substitution by setting \( y \) back to \( s^2 \) and solve for \( s \) to get the final solutions.

It's important to check that the substitutions made are reversible and that they accurately represent solutions for the original variable.

In algebra, it's critical to differentiate between real and complex solutions. A polynomial equation may have a number of solutions that include both real numbers and complex numbers (that include the imaginary unit \( i \) ). However, for many practical applications, including the exercise at hand, we are often interested only in real solutions.

For a real solution, the variable squared must result in a non-negative number since the square root of a negative number is not a real number. In the case of the exercise \( 4s^4 + 11s^2 - 3 = 0 \), we see that after applying the quadratic formula and substitution method, one of the solutions for \( s^2 \) is negative (\( s^2 = -3.463 \) ), which does not give a real value when we take the square root. Hence, we discard this solution.

On the other hand, the positive \( s^2 = 0.213 \) does have a square root that is real, resulting in two real solutions once we account for both the positive and negative square roots. As a good practice, these found solutions should always be verified by substituting back into the original equation to ensure they are true solutions.