91Ó°ÊÓ

Fractional exponents are a unique way to express power and roots of numbers. When we say something like \( x^{\frac{2}{3}} \), it actually represents a root combined with an exponentiation.
So, when you see \( x^{\frac{2}{3}} \), here's what's happening:
- The fraction \( \frac{2}{3} \) implies a cube root (because of the denominator 3) and squaring (because of the numerator 2).
- This means \( x^{\frac{2}{3}} = (x^2)^{\frac{1}{3}} \) or \( (x^{\frac{1}{3}})^2 \).
This kind of expression is useful when solving equations because, sometimes, substituting complicated root-exponent expressions can simplify problems enormously. Adopting substitution for these can transform a seemingly complex problem into a simpler form, like a quadratic equation.
Understanding and working with fractional exponents is a vital skill in algebra, allowing complex transformations that wouldn't otherwise be feasible.

The quadratic formula is a powerful tool that provides solutions to quadratic equations. A quadratic equation takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
To find the roots of the quadratic equation, we use the formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here's a step-by-step breakdown of how to apply it:

• Calculate the discriminant \( b^2 - 4ac \). This part of the formula determines how many solutions the equation has.
• Plug the values of \( a \), \( b \), and \( c \) into the quadratic formula.
• Compute the square root of the discriminant, add and subtract it from \( -b \), then divide by \( 2a \) to find the values of \( y \).

In this problem, using the quadratic formula transformed a fractional exponent equation into a solvable quadratic equation. Mastery of the quadratic formula helps in swiftly tackling a broad variety of challenges in mathematics.

The substitution method is a strategic algebraic technique where you replace variable expressions with simpler placeholders, often easing the complexity of a problem.
For example, in dealing with the original equation like the one given here, using the substitution \( y = x^{\frac{1}{3}} \) transforms a complex expression into a more familiar quadratic form: \( 6y^2 - 5y - 6 = 0 \).
This process simplifies the mathematical work required. Here’s how it works:

• Identify expressions in the equation that seem overly complicated.
• Define them in terms of new variables like \( y \).
• Rewrite the entire equation using these new terms.
• Solve the simpler, converted equation.

The substitution method is especially useful when working with polynomials and equations involving radical exponents. It helps convert challenging problems into ones that can be approached using basic algebraic techniques.

In mathematics, verifying solutions confirms the accuracy of your answer by substituting it back into the original equation. It acts as a safeguard against mistakes.
Here's how you verify solutions:

• Take the solution you calculated and substitute it back into the original equation.
• Evaluate to see if the left-hand side equals the right-hand side.
• If both sides equal, your solution is confirmed. If not, revisit your calculations.

In the given exercise, both potential solutions of \( x \) derived from distinct solutions of \( y \) should hold true in the original equation \( 6 x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} - 6 = 0 \).
This step won’t just affirm that the work is correct; it also builds mathematical intuition by reinforcing the conceptual connections between substitute variables and their roles in the equation. Recognizing the importance of solution verification leads to more robust problem-solving skills.