Problem 45 - Algebra and Trigonometry

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. $$x-13 \sqrt{x}+40=0$$

Step-by-Step Solution

Answer

The solutions for the equation \(x-13 \sqrt{x}+40=0\) are \( x = 64 \) and \( x = 25 \)

Step 1 : Variable change

Let's denoted by\$ u \$ the square root of \$ x \$, that is \$ u = \sqrt{x} \$. Or equivalently, \$ x = u^2 \$. Replacing \$ x \$ with \$ u \$ in the equation we get a quadratic equation in \$ u \$: \( u^2 - 13u + 40 = 0 \$.

Step 2 : Solve the quadratic equation

This is a standard quadratic equation and can be solved by factoring: \( (u - 8)(u - 5) = 0 \$. Setting each factor equals zero gives solutions\$ u = 8 \$ or\$ u = 5 \$.

Step 3 : Back substitution

Recall that \$ u = \sqrt{x} \$. Replacing \$ u \$ by \$ \sqrt{x} \$ gives the solutions of original equation: \$ x = 64 \$ for \$ u = 8 \$ and \$ x = 25 \$ for \$ u = 5 \$.

Key Concepts

Variable Substitution

When faced with complex equations, one method to simplify them is by using variable substitution. This technique involves replacing a variable or expression with another variable, often simplifying the equation into a more manageable form.
In the given equation, we start with \( x - 13 \sqrt{x} + 40 = 0 \). Since the expression involves \( \sqrt{x} \), we can use this as our substitution point. Let's introduce a new variable \( u \) such that \( u = \sqrt{x} \).
With this substitution, the equation can be rewritten as \( u^2 - 13u + 40 = 0 \), which is now a quadratic equation in terms of \( u \). This makes the solution process much simpler and more straightforward, allowing us to use standard methods for solving quadratic equations.

Factoring

Factoring is a key technique used to solve quadratic equations efficiently. The idea is to express the quadratic equation as a product of two binomials. For our equation, after substitution: \( u^2 - 13u + 40 = 0 \), we need to look for two numbers that multiply to 40 and add up to -13.
These numbers are -8 and -5. Therefore, the equation factors to \((u - 8)(u - 5) = 0\).
This factored form allows us to identify the possible solutions quickly by setting each factor equal to zero.

Square Root

In the context of this problem, recognizing and working with square roots is essential. Initially, we performed variable substitution with \( u = \sqrt{x} \). The operation of squaring and finding square roots is key here, as it let us convert the original equation into a simpler quadratic form.
The square root, denoted as \( \sqrt{a} \), represents a value which, when multiplied by itself, gives \( a \). Hence, when we back substitute \( u = 8 \) and \( u = 5 \) into \( x = u^2 \), we take advantage of the property function of square roots.
This relationship is crucial for verifying the solutions we acquire from the quadratic equation, transforming \( u = \sqrt{x} \) back to \( x \) itself.

Back Substitution

Once the quadratic equation is solved, the solutions we find are in terms of the substituted variable, \( u \). To find the solutions to the original equation, back substitution is necessary. This process involves reverting to the original variable in terms of the solutions obtained.
From our factored equation, we found solutions \( u = 8 \) and \( u = 5 \). Recall that \( u = \sqrt{x} \). We substitute these values back to determine \( x \):

• If \( u = 8 \), then \( \sqrt{x} = 8 \), leading to \( x = 8^2 = 64 \).
• If \( u = 5 \), then \( \sqrt{x} = 5 \), leading to \( x = 5^2 = 25 \).

This method ensures that we correctly solve for the values of \( x \) that satisfy the original equation.