Problem 75 - Intermediate Algebra

Question

Solve each equation. $$ x+\sqrt{x}-2=0 $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \).

Step 1: Isolate the Square Root

Start by moving the non-square root terms to one side of the equation. Given the equation,\[ x + \sqrt{x} - 2 = 0 \]we can add 2 to both sides to get:\[ x + \sqrt{x} = 2 \]

Step 2: Substitution Strategy

Introduce a substitution to simplify the process. Let \( y = \sqrt{x} \). Then \( y^2 = x \). Substitute this into \( x + \sqrt{x} = 2 \) to give:\[ y^2 + y = 2 \]

Step 3: Solve the Quadratic Equation

Rearrange the quadratic equation derived from the substitution:\[ y^2 + y - 2 = 0 \]This is a standard quadratic equation and can be solved using factoring or the quadratic formula. Factoring yields:\[ (y + 2)(y - 1) = 0 \]Thus, \( y = -2 \) or \( y = 1 \).

Step 4: Back Substitution

Since \( y = \sqrt{x} \), and the square root can only be non-negative, we disregard \( y = -2 \). So, we have \( y = 1 \), which gives:\[ \sqrt{x} = 1 \]

Step 5: Solve for x

Square both sides to solve for \( x \):\[ (\sqrt{x})^2 = 1^2 \]thus,\[ x = 1 \]

Step 6: Verify the Solution

Substitute \( x = 1 \) back into the original equation to ensure it satisfies the equation. The original equation is:\[ x + \sqrt{x} - 2 = 0 \]Substitute \( x = 1 \):\[ 1 + \sqrt{1} - 2 = 0 \]\[ 1 + 1 - 2 = 0 \]\[ 0 = 0 \]The equation holds true, confirming \( x = 1 \) is a valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are an essential concept in intermediate algebra that usually appear in the form \( ax^2 + bx + c = 0 \). These equations can be solved using various methods, such as factoring, completing the square, or the quadratic formula.
In this exercise, after using a substitution involving the original problem \( x + \sqrt{x} - 2 = 0 \), a quadratic equation is formed: \( y^2 + y - 2 = 0 \).
This is a standard quadratic equation where:

• \( a = 1 \)
• \( b = 1 \)
• \( c = -2 \)

Factoring is often the most straightforward way to solve simple quadratic equations, especially when the coefficients are small and manageable. In this instance, the quadratic equation \( y^2 + y - 2 = 0 \) can be factored into \((y + 2)(y - 1) = 0\), yielding solutions \( y = -2 \) and \( y = 1 \). However, only solutions that respect the initial conditions will finalize the solution.

Square Roots

In mathematics, a square root of a number is a value that, when multiplied by itself, gives the original number. The symbol for a square root is \( \sqrt{} \). Understanding square roots is crucial, as they appear frequently in algebraic problems.
Within the given exercise \( x + \sqrt{x} - 2 = 0 \), \( \sqrt{x} \) indicates the square root of \( x \), which is pivotal in the substitution step.
It's important to note that square roots are non-negative when dealing with real numbers. This means that a square root like \( \sqrt{x} \) will always result in a value that is zero or positive.

• When solving \( \sqrt{x} = 1 \), you square both sides to eliminate the square root: \( (\sqrt{x})^2 = 1^2 \), resulting in \( x = 1 \).
• By ensuring non-negative results, we focus only on valid mathematical solutions. For instance, \( y = \sqrt{x} \) must give non-negative \( y \).

Therefore, when you encounter square roots in equations, verify your solutions satisfy all mathematical properties associated with square roots.

Substitution Method

The substitution method is a powerful technique to simplify complex equations, making them easier to solve. By substituting part of the expression with a different variable, it creates a simpler form.
For example, in the equation \( x + \sqrt{x} - 2 = 0 \), substituting \( y = \sqrt{x} \) transforms the equation into \( y^2 + y = 2 \). This substitution allows the equation to be approached via solving a quadratic equation rather than manipulated as a more complex original form.
Benefits of substitution include:

• Simplifying equations by reducing complexity.
• Facilitating the use of different algebraic strategies after substitution.
• Allowing focus on easier solutions which can be back-substituted.

The primary goal is to make the equation more manageable and straightforward. After solving \( y \), reverse the substitution to find \( x \), the original variable. Always verify that the solution satisfies the initial conditions to ensure its validity.