Problem 19 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ \sqrt{7-x}=x-5 $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = 6 \).

Step 1: Understand the Equation

We need to solve the equation \( \sqrt{7-x} = x - 5 \). This equation involves a square root and a linear term on the right side. Our goal is to find the value of \( x \) that satisfies this equation.

Step 2: Remove the Square Root

To eliminate the square root, we square both sides of the equation. This gives us:\[ (\sqrt{7-x})^2 = (x-5)^2 \]Simplifying, we get:\[ 7-x = (x-5)^2 \]

Step 3: Expand the Quadratic

Expand the right side of the equation:\[ x^2 - 10x + 25 \] This changes the equation to:\[ 7 - x = x^2 - 10x + 25 \]

Step 4: Rearrange the Terms

Move all terms to one side of the equation to set it to zero:\[ 0 = x^2 - 10x + 25 + x - 7 \]Combine like terms:\[ 0 = x^2 - 9x + 18 \]

Step 5: Factor the Quadratic

Next, we factor the quadratic equation:\[ x^2 - 9x + 18 = 0 \]The factors of 18 that add to -9 are -3 and -6, so we factor it as:\[ (x-3)(x-6) = 0 \]

Step 6: Find the Solutions

Set each factor equal to zero to solve for \( x \):1. \( x - 3 = 0 \) gives \( x = 3 \)2. \( x - 6 = 0 \) gives \( x = 6 \)Thus, we have two potential solutions: \( x = 3 \) and \( x = 6 \).

Step 7: Check the Solutions

We need to verify which solutions satisfy the original equation.For \( x = 3 \), substitute into the original equation:\[ \sqrt{7-3} = 3 - 5 \]\[ 2 eq -2 \] (not valid)For \( x = 6 \), substitute into the original equation:\[ \sqrt{7-6} = 6 - 5 \]\[ 1 = 1 \] (valid)Thus, the solution is \( x = 6 \).

Key Concepts

Square Root Equations

Square root equations involve variables under a square root symbol. To solve these equations, the first step is to remove the square root. This is typically done to simplify the equation and isolate the variable.
To remove a square root:

• Square both sides of the equation.
• This should eliminate the square root on one side of the equation.

Once the square root is removed, the problem often becomes easier or transforms into another type of equation, such as a quadratic equation. However, be cautious because squaring both sides can sometimes introduce extraneous solutions.
These are solutions that may fit the squared equation but not the original problem. It's important to substitute the solutions back into the original equation to verify their validity. This step ensures any solutions that do not actually satisfy the equation are discarded.

Quadratic Equations

Quadratic equations are fundamental in algebra and involve variables raised to the power of two, typically written in the standard form: \[ ax^2 + bx + c = 0 \]Here, "\( a \)," "\( b \)," and "\( c \)" are constants, with "\( a \)" not equal to zero. Solving quadratic equations means finding the values of \( x \) which satisfy the equation. There are several methods for solving them:

• Factoring: If the quadratic is factorable, it is often the simplest method. Identify factors of the constant term "\( c \)" that add up to the coefficient "\( b \)."
• Quadratic formula: Given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] this method is universal but may be more complex.
• Completing the square: A less common method where a perfect square trinomial is created from the quadratic equation, often used to derive the quadratic formula itself.

Factoring is particularly efficient for simple quadratics, as seen in the above solution where we factorized the equation after rearranging terms. This often provides a quick solution if factoring is straightforward.

Factoring Quadratics

Factoring quadratic equations involves expressing the quadratic polynomial in the form of two binomial expressions. This is a highly useful technique in both solving quadratic equations and simplifying algebraic expressions. To factor a quadratic like \( ax^2 + bx + c \), follows these steps:

• Identify the product of "\( a \)" and "\( c \)." Find two numbers that multiply to this product and add to "\( b \)."
• Use these numbers to split the middle term, "\( bx \)," and rearrange the expression.
• Factor by grouping, which involves grouping terms that share common factors.
• Confirm that factoring is correct by expanding the binomials to check they multiply back to the original quadratic expression.

In our solution, the quadratic \[ x^2 - 9x + 18 \]was factored into \[(x-3)(x-6),\] by finding numbers \(-3 \text{ and } -6,\) that multiply to give \(18\) and add up to \(-9,\) thus making it easier to identify solutions.