Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ \sqrt{3-x}-x=3 $$

Step-by-Step Solution

Answer

The solutions are \(x = -1\) and \(x = -6\).

Step 1: Isolate the Square Root

Start by isolating the square root on one side of the equation. We have \( \sqrt{3-x} - x = 3 \). Add \( x \) to both sides to get \( \sqrt{3-x} = x + 3 \).

Step 2: Eliminate the Square Root

Square both sides of the equation to eliminate the square root. This yields \((\sqrt{3-x})^2 = (x+3)^2 \), simplifying to \(3-x = x^2 + 6x + 9 \).

Step 3: Rearrange into a Quadratic Equation

Rearrange the equation by bringing all terms to one side to form a quadratic equation: \(x^2 + 6x + 9 + x - 3 = 0 \). Simplify to get \(x^2 + 7x + 6 = 0 \).

Step 4: Solve the Quadratic Equation

Factor the quadratic equation \(x^2 + 7x + 6 = 0 \). Look for two numbers that multiply to 6 and add up to 7. These numbers are 1 and 6, giving the factors \((x+1)(x+6) = 0\).

Step 5: Find the Solutions

Set each factor equal to zero to find \(x\): \(x + 1 = 0\) or \(x + 6 = 0\). Solving these gives \(x = -1\) and \(x = -6\).

Step 6: Verify the Solutions

Substitute \(x = -1\) and \(x = -6\) back into the original equation to verify they are true solutions. For \(x = -1\), \(\sqrt{3-(-1)} - (-1) = 3\) holds as true. For \(x = -6\), \(\sqrt{3-(-6)} - (-6) = 3\) also holds as true. Both solutions are valid.

Key Concepts

Understanding Algebra

Algebra involves the manipulation of mathematical symbols to solve equations or understand relationships between variables. It is a fundamental part of mathematics, dealing with expressions, equations, and the rules for manipulating these symbols to discover unknown values.
Algebra can be seen as a generalized form of arithmetic. In solving the given equation \( \sqrt{3-x} - x = 3 \), algebra helps us rearrange and simplify the equation to make it easier to solve.
In our original equation, invoking algebra let us isolate the square root by performing operations on both sides, such as adding \( x \) to each side to get \( \sqrt{3-x} = x + 3 \). This set the stage for solving the equation.

Square Root and Its Properties

A square root asks the question: "What number, when multiplied by itself, will yield the given number?" Understanding square roots is essential for solving equations involving radical expressions. It's a common component of algebraic equations like our example.
When we isolated \( \sqrt{3-x} \) in the equation, we moved to eliminate the square root so the expression became easier to handle. We did this by squaring both sides, which is a typical algebraic operation to remove square roots.
This action was supported by the property that squaring a square root returns the radicand (inside value), turning our equation from \( \sqrt{3-x} = x+3 \) into \( 3-x = (x+3)^2 \). This allowed us to progress to a more conventional quadratic equation.

Factoring Quadratic Equations

Factoring is the process of breaking down expressions into simpler terms ("factors") that, when multiplied together, give the original expression. It's a crucial skill in solving quadratic equations.
In our exercise, once the equation was rearranged to \( x^2 + 7x + 6 = 0 \), factoring came into play. The task is to find two numbers that multiply to give the constant term (6) and add up to the coefficient of the linear term (7).
In this case, the numbers are 1 and 6, which allowed us to factor the equation as \((x+1)(x+6)=0\). Factoring simplifies the equation into a product of binomials, enabling us to find the solutions by setting each product to zero.

Verification of Solutions

Once potential solutions are found, it is essential to verify them within the context of the original equation. Verification ensures that the solutions are valid and satisfy the initial condition given in the problem.
For our equation, after factoring, we found potential solutions \(x = -1\) and \(x = -6\). Verification requires substituting these values back into the original equation \( \sqrt{3-x} - x = 3 \) to check their validity.
For \(x = -1\), the left side becomes \( \sqrt{3 - (-1)} + 1 = 3\), which simplifies to 3, matching the right side of the equation. Similarly, substituting \(x = -6\) results in an equivalent expression, confirming both are correct solutions.