Problem 73 - Elementary Algebra

Question

Solve and round off the solutions to the nearest hundredth. $$ (x+3)(x-7)=11-4 x $$

Step-by-Step Solution

Answer

The solutions are approximately \(x \approx 5.66\) and \(x \approx -5.66\).

Step 1: Expand the Left Side

We start by expanding the left side of the equation \[(x+3)(x-7) \]Using the distributive property:\[x(x-7) + 3(x-7) = x^2 - 7x + 3x - 21\]Thus, it simplifies to \[x^2 - 4x - 21\]

Step 2: Set the Equation to Zero

Next, we rearrange the equation to bring all terms to one side to equal zero:\[x^2 - 4x - 21 = 11 - 4x\]Subtract \((11 - 4x)\) from both sides:\[x^2 - 4x - 21 - 11 + 4x = 0\]This simplifies to:\[x^2 - 32 = 0\]

Step 3: Solve the Equation

Now, solve the equation \[x^2 - 32 = 0\]Add 32 to both sides:\[x^2 = 32\]Take the square root of both sides:\[x = \pm \sqrt{32}\]

Step 4: Simplify the Square Root

Simplifying the square root gives:\[x = \pm \sqrt{16\times 2} = \pm 4\sqrt{2}\]Now, approximate \(\sqrt{2}\) as 1.414 and calculate:\[x \approx \pm 4 \times 1.414 = \pm 5.656\]

Step 5: Round the Solution

Round \(\pm 5.656\) to the nearest hundredth to get:\[x \approx 5.66\] and \[x \approx -5.66\]

Key Concepts

Distributive Property

In any algebraic expression, the distributive property allows you to simplify expressions by spreading the multiplication over addition or subtraction within parentheses. This property is written as:

• \( a(b + c) = ab + ac \)
• \( a(b - c) = ab - ac \)

In our original exercise, we tackled \( (x+3)(x-7) \). To use the distributive property, each term inside the first parenthesis multiplies with each term in the second. For example:- The term \( x \) multiplies with \( x \) and \( -7 \), resulting in \( x^2 - 7x \).- The term \( 3 \) then multiplies with \( x \) and \( -7 \), giving \( 3x - 21 \).By combining these, the expression simplifies to \( x^2 - 4x - 21 \).It's key to distribute carefully and ensure each multiplication is dealt with properly, bringing accuracy to the equation's simplification process.Every time you are faced with multiplication over addition or subtraction, remember this property. It helps open up complex expressions into more manageable parts.

Square Root

A square root is essentially the opposite of squaring a number. Finding the square root of a number involves determining which number, when multiplied by itself, gives you the original number. Mathematically, it's expressed as:- \( \sqrt{a} \) where \( a\) is the number you want the square root of.In our exercise, we ended up with \( x^2 = 32 \). To solve for \( x \), we take the square root of both sides:- \( x = \pm \sqrt{32} \).The \( \pm \) symbol, read as "plus-minus," is essential here. It indicates there are two possible solutions, one positive and one negative.Then, simplifying \( \sqrt{32} \) involves finding factors of 32 that are perfect squares. We see that:- \( 32 = 16 \times 2 \), with 16 being a perfect square.- Therefore, \( \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \).This simplification allows for easier calculations, especially when needing an approximate decimal value.

Simplifying Algebraic Expressions

After expanding and using the distributive property in our given problem, we often need to simplify further to work with a cleaner expression. When simplifying, we aim to combine like terms and perform arithmetic operations that help tidy up the expression.For example, from the expansion in our problem, we had:- \( x^2 - 4x - 21 = 11 - 4x \).By moving all terms to one side, the expression simplifies:- Subtract \( (11 - 4x) \) from both sides to get:- \( x^2 - 4x - 21 - 11 + 4x = 0 \).Here, like terms \( -4x \) and \( +4x \) cancel each other out, and we combine constant terms:- \( x^2 - 32 = 0 \).In such simplifications, always check for terms that can be combined and look for operations that remove unnecessary components, making the problem-solving process more efficient.

Solving Equations

Solving equations often requires carefully thought-out steps and operations that gradually isolate the variable, giving us solutions. In our quadratic equation problem, after simplification, we arrive at:- \( x^2 = 32 \).We solve for \( x \) by applying algebraic techniques. First, let's take these steps:

• Add or subtract terms to isolate the variable term: make sure the \( x^2 \) term stands alone.
• Apply operations, like taking square roots, to solve for the variable.

Once we've done that:- We take the square root of both sides, yielding \( x = \pm \sqrt{32} \).Finally, it's often necessary to approximate or round solutions, especially when the problem requires a decimal approximation. Here, after simplifying \( \sqrt{32} \) as \( 4 \sqrt{2} \) and calculating an approximate value, we reach:- \( x \approx 5.66 \) and \( x \approx -5.66 \).Rounding to the nearest hundredth is one way to make solutions more digestible and applicable, especially in real-world contexts where precision is required. Always ensure, when you solve for an equation, to verify each step for accuracy and couple it with the necessary rounding when needed.