91Ó°ÊÓ

The substitution method is a powerful technique in algebra used to simplify complex equations by introducing a new variable. In this exercise, we used a substitution to reduce an equation with a fractional exponent into a more familiar form. We defined a new variable:

• Let \( y = x^{1/3} \).

This decision transforms the original equation \( x^{2/3} - 2x^{1/3} - 15 = 0 \) into a quadratic equation \( y^2 - 2y - 15 = 0 \). Such substitutions often make it easier to apply standard algebraic methods, like solving quadratic equations, allowing us to solve for the new variable before reverting back to the original variable. Remember, the key here is to choose the appropriate substitution that simplifies the equation. This step is crucial because it serves as the bridge between a complex form and a simpler, more solvable form.

A quadratic equation is a polynomial equation of the second degree, typically expressed as \( ax^2 + bx + c = 0 \). In this exercise, after substituting \( y = x^{1/3} \), we ended up with a quadratic equation in the form of \( y^2 - 2y - 15 = 0 \). Quadratic equations can be solved using several methods:

• Factoring
• Using the quadratic formula
• Completing the square

In our case, we used factoring because we could easily find two numbers that multiply to \(-15\) and add to \(-2\). These numbers were \(3\) and \(-5\), resulting in the factored form \((y - 5)(y + 3) = 0\). Factoring is a quick and efficient method particularly when the equation is factorable with integers. But it is essential to explore other methods for more complex equations where factoring may not be straightforward.

Finding the roots or solutions of a polynomial is a fundamental aspect of solving polynomial equations. After rewriting the original equation as a quadratic \( y^2 - 2y - 15 = 0 \) and factoring it into \((y - 5)(y + 3) = 0\), we determine the roots by setting each factor equal to zero:

• \( y - 5 = 0 \), yielding \( y = 5 \).
• \( y + 3 = 0 \), yielding \( y = -3 \).

These are known as the 'roots' or 'solutions' of the quadratic equation. Each root represents a solution to the equation that satisfies the condition of making the equation equal zero. Once the quadratic roots are found, the process involves reversing the substitution to find the roots of the original equation with respect to \( x \). Understanding how to find polynomial roots allows you to solve a wide array of problems across different branches of mathematics.

Verification of solutions is an important step to ensure the solutions found satisfy the original equation. In our example, after solving for \( y \) and substituting back to find \( x \), it is crucial to substitute these solutions back into the original equation to verify them. For \( x = 125 \):

• Substituting back, \( 125^{2/3} - 2\times125^{1/3} - 15 = 0 \).
• Simplifies to \( 25 - 10 - 15 = 0 \), which is true.

For \( x = -27 \):

• \((-27)^{2/3} - 2\left((-27)^{1/3}\right) - 15 = 0 \).
• This simplifies to \( 9 + 6 - 15 = 0 \), which is also true.

Verification confirms that the solutions are indeed correct and satisfy the original equation, ensuring the accuracy of the answers provided. It is a good practice to always verify your solutions to avoid errors and misunderstandings.