Problem 45 - Brief Applied Calculus

Question

Solve each equation by factoring or the Quadratic Formula, as appropriate. $$ 4 x^{2}+24 x+40=4 $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = -3 \).

Step 1: Move All Terms to One Side of the Equation

First, write the equation in the standard form of a quadratic equation: \[ 4x^2 + 24x + 40 = 4 \]Subtract 4 from both sides:\[ 4x^2 + 24x + 36 = 0 \]

Step 2: Factor Out the Greatest Common Factor

Notice that each term in the equation \(4x^2 + 24x + 36\) can be divided by 4. Factor out the greatest common factor (GCF), which is 4:\[ 4(x^2 + 6x + 9) = 0 \]

Step 3: Factor the Quadratic Expression

Now, factor the quadratic expression \(x^2 + 6x + 9\). Find two numbers that multiply to 9 and add up to 6, which are 3 and 3:\[ 4(x+3)(x+3) = 0 \]This can be written as:\[ 4(x+3)^2 = 0 \]

Step 4: Solve for x by Setting the Factor to Zero

Instead of dealing with the constant 4, set the expression \((x+3)^2\) equal to zero:\[ (x + 3)^2 = 0 \]Take the square root of both sides:\[ x + 3 = 0 \]

Step 5: Solve for x

Solve the equation \(x + 3 = 0\) for \(x\):\[ x = -3 \]

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial method for solving quadratic equations. It involves breaking down an expression into products of simpler factors. For example, in the equation given, after moving all terms to one side, we have the expression \(4x^2 + 24x + 36 = 0\). The first step is to identify any common factors in the polynomial. By recognizing that each term in the trinomial can be divided by 4, we factor out this greatest common factor (GCF). This simplifies our expression to \(4(x^2 + 6x + 9) = 0\).
You can think of factoring like reversing multiplication: you are finding terms that when multiplied together give you the original polynomial.
Once you have factored out the GCF, the next step is to factor the resulting quadratic, \(x^2 + 6x + 9\), into a product of two binomials. Here, it factors neatly into \((x+3)(x+3)\) because we find numbers 3 and 3, which multiply to 9 and add up to 6.
By writing it as \((x+3)^2\), we simplify the solving process further. Remember, factoring transforms a polynomial into a product of simpler expressions that can be set to zero to find solutions.

Quadratic Formula

The quadratic formula offers a reliable way to solve any quadratic equation, especially when factoring seems complex or impossible. The formula is: \[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]where \(a\), \(b\), and \(c\) are coefficients from the equation \(ax^2 + bx + c = 0\).
This formula can solve any quadratic equation, even when the numbers involved are not easy integers, as with \(x^2 + 6x + 9\). However, in our example, we took a simpler approach with factoring, because the equation factored neatly.
The usefulness of the quadratic formula lies in its versatility:

• It provides solutions directly when factoring is difficult.
• It reveals complex roots through the discriminant \(b^2 - 4ac\).
• The formula helps understand if any solutions exist based on the discriminant value.

So, while we did not need it here due to the simplicity of our factors, know it is a powerful tool in your toolkit for solving quadratic equations.

Solving Equations

Solving quadratic equations can be achieved through various methods, each offering a unique approach to find the unknown \(x\). Once you have either factored the quadratic expression or applied the quadratic formula, you set the factors equal to zero to find the solutions. In our provided example, we factored \((x+3)^2 = 0\).
Setting \((x+3)^2 = 0\) implies that \((x+3) = 0\). This gives us a solution:

• Move the constant 3 to the opposite side by subtracting 3 from both sides.
• The result is: \(x = -3\).

Each method of solving has its own advantages. Factoring, as demonstrated, allows for easy identification of simple solutions. Using the quadratic formula opens avenues to more complicated equations. Consider which method suits the given problem the best for a more efficient solution.