Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Solve: } x-5 \sqrt{x}+6=0 $$

Short answer

The solutions are \( x = 4 \) and \( x = 9 \).

Step-by-step solution

Introduce a new variable

Let us introduce a new variable to simplify the equation. Let \[ y = \sqrt{x} \]. Then, squaring both sides gives \[ y^{2} = x \].

Substitute the new variable

Substitute \( y \) and \( y^2 \) into the original equation: \[ x - 5 \sqrt{x} + 6 = 0 \] becomes \[ y^{2} - 5y + 6 = 0 \].

Solve the quadratic equation

Solve the quadratic equation \[ y^{2} - 5y + 6 = 0 \]. This can be factored as \[ (y - 2)(y - 3) = 0 \].

Find the values of y

Set each factor equal to zero: \[ y - 2 = 0 \Rightarrow y = 2 \] and \[ y - 3 = 0 \Rightarrow y = 3 \].

Back-substitute the values of y

Recall that \( y = \sqrt{x} \). So we back-substitute to find \[ \sqrt{x} = 2 \Rightarrow x = 4 \] and \[ \sqrt{x} = 3 \Rightarrow x = 9 \].

Verify the solutions

Substitute \( x = 4 \) and \( x = 9 \) back into the original equation to verify:For \( x = 4 \): \[ 4 - 5 \sqrt{4} + 6 = 4 - 10 + 6 = 0 \Rightarrow \text{True} \]For \( x = 9 \): \[ 9 - 5 \sqrt{9} + 6 = 9 - 15 + 6 = 0 \Rightarrow \text{True} \].Therefore, both solutions are correct.

Key concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). Here, \(a, b,\) and \(c\) are constants, and \(x\) represents the variable. Quadratic equations are significant in various fields, such as physics, engineering, and finance. In this problem, after substituting \(y = \sqrt{x}\), we transformed the given equation into a quadratic equation: \[ y^2 - 5y + 6 = 0 \].
To solve quadratic equations, you can use several methods:

• Factoring: Expressing the quadratic equation as a product of its linear factors.
• Quadratic Formula: Using the formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] to find the roots.
• Completing the Square: Rearranging the quadratic equation to form a perfect square trinomial.
  

Understanding how to solve quadratic equations is crucial, as it opens up pathways to more advanced topics in algebra and calculus.

Substitution Method

The substitution method is a technique used to solve equations by introducing a new variable to simplify the problem. In our exercise, we used \( y = \sqrt{x} \), and then squared both sides to get \( y^2 = x \). This transformed the original equation into a quadratic equation in terms of \( y \).
This method is beneficial when dealing with equations involving square roots or other complex expressions, as it:

• Reduces the complexity of the equation.
• Allows us to apply well-known solving techniques, such as factoring for quadratic equations.
  

Once you've solved for the new variable, don't forget to back-substitute to find the value of the original variable. This is a crucial step to ensure your solution is complete.

Square Root

The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). It's denoted as \( \sqrt{x} \). Positive real numbers have two square roots: a positive and a negative root, but we mostly consider the principal (positive) square root. In our problem, after substituting \( y = \sqrt{x} \) and solving the quadratic equation, we found \( y = 2 \) and \( y = 3 \).
We then back-substituted to find:

• \( \sqrt{x} = 2 \implies x = 4 \)
• \( \sqrt{x} = 3 \implies x = 9 \)

Always verify your solutions by substituting them back into the original equation. This ensures they satisfy the equation, confirming your answers are correct.