Problem 46 - College Algebra

Question

Solve each equation by making an appropriate substitution. $$2 x-7 \sqrt{x}-30-0$$

Step-by-Step Solution

Answer

The solution steps are first to determine an appropriate substitution, which in this case is \(y = \sqrt{x}\). This transforms the original equation \(2x-7 \sqrt{x}-30 = 0\) into a quadratic format: \(2y^2 - 7y - 30 = 0\). Solving this quadratic equation will produce the values for \(y\), these are then substituted back into \(y = \sqrt{x}\) to find the corresponding values for \(x\). Lastly, it is important to check these answers against the original equation to ensure that they are valid solutions, that is they are not extraneous solutions.

Step 1: Identify the suitable substitution

Given the equation \(2x-7 \sqrt{x}-30 = 0\), you can see that if a new variable is used to take the place of the square root of x, it will transform the equation into a simpler quadratic form. Let's substitute \(y = \sqrt{x}\) or \(y^2 = x\). The original equation then changes to a form which is easier to solve.

Step 2: Substitute and simplify

Replacing \(x\) in the original equation with \(y^2\), we get \(2y^2 - 7y - 30 = 0\). Now this equation looks simpler to solve as it turns into a standard quadratic equation with y being the variable.

Step 3: Solve the quadratic equation

Solving the quadratic equation, we can be use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In our equation, \(a = 2\), \(b = -7\), and \(c = -30\). Plugging these values into the quadratic formula gives the solutions for \(y\).

Step 4: Substitute back to find x

Remember, we initially substituted \(y = \sqrt{x}\). So, now we substitute the found values for \(y\) back to this equation to find the values for \(x\)

Step 5: Discard any extraneous solutions

It's important to remember that the process of squaring during the solution process often introduces extraneous solutions. So always check the answers into the original equation to ensure they satisfy it.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation that generally takes the form \(ax^2 + bx + c = 0\). These equations are called "quadratic" because they involve variables, such as \(x\), raised to the power of two. They can appear in various scientific and mathematical applications and usually involve unknown values for which we need to solve. In our original exercise, by making a substitution \(y = \sqrt{x}\), we are transforming our equation into the quadratic form, \(2y^2 - 7y - 30 = 0\). This transformation simplifies problem-solving greatly, turning a complex equation into one that fits the simpler structure of a quadratic equation.

Solving Equations

Solving equations involves finding the values of variables that make an equation true. When working with equations, we use various techniques to isolate variables and solve for them. For quadratic equations, like in Step 3 of the solution, we often use the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula allows us to solve any quadratic equation by plugging in values for \(a\), \(b\), and \(c\) from the equation format \(ax^2 + bx + c = 0\). In more straightforward scenarios, factoring or completing the square might be other methods employed.
Regardless of the method, the goal remains clear – find the unknown values that satisfy our equation. After using these techniques, always verify that your solutions work by substituting them back into the original equation.

Square Root Substitution

Square root substitution is a clever technique used to simplify equations, especially when dealing with terms involving \(\sqrt{x}\). By letting \(y = \sqrt{x}\), we effectively remove the square root operation, enabling us to work with polynomial terms. This method was employed in our problem where \(\sqrt{x}\) was present.

• First, we identified that substituting \(y = \sqrt{x}\) would help simplify the equation.
• The substitution transformed the original equation to a familiar quadratic form.

Using square root substitution not only simplifies equations but also opens a path to using powerful polynomial-solving techniques like the quadratic formula. Once substituted, the problem becomes a polynomial equation, vastly simplifying our approach and solution.

Extraneous Solutions

Extraneous solutions often appear as an unintended consequence of equation manipulation. They are solutions that emerge from the mathematics but do not satisfy the original problem.

• Such solutions occur most commonly when using methods like substitution or squaring.
• In our problem, after substituting back from \(y\) to \(x\), it's crucial to validate solutions by plugging them back into the original equation \(2x - 7\sqrt{x} - 30 = 0\).

Detection and rejection of extraneous solutions safeguard the accuracy of problem-solving. Always check every derived solution against the initial equation to confirm it is valid.