Problem 61 - College Algebra

Question

Solve each equation in Exercises \(55-64\) using the quadratic formula. $$ 4 x^{2}=2 x+7 $$

Step-by-Step Solution

Answer

The solutions to the quadratic equation are \(x = \frac{2 + \sqrt{116}}{8}\) and \(x = \frac{2 - \sqrt{116}}{8}\)

Step 1: Rewrite the equation in standard form

Firstly, the given equation \(4x^{2} = 2x + 7\) is rewritten in standard form. The standard form of a quadratic equation is \( ax^{2} + bx + c = 0 \). So rewrite the equation as: \(4x^{2} - 2x - 7 = 0\)

Step 2: Identify a, b and c

The coefficients of the equation are matched with the standard form to identify the variables a, b, and c. Thus, \(a = 4\), \(b = -2\), and \(c = -7 \)

Step 3: Substitution into the quadratic formula

Now the values are substituted into the quadratic formula, to find the value of x: \(x = \frac{-(-2) \pm \sqrt{(-2)^{2}-4*4*(-7)}}{2*4}\), which simplifies to: \(x = \frac{2 \pm \sqrt{4+112}}{8}\)

Step 4: Simplifying the formula

The numbers within the square root are combined and simplified, gives: \(x = \frac{2 \pm \sqrt{116}}{8}\)

Step 5: Finding the solution

After reducing the square root and simplifying, the solutions are: \(x = \frac{2 + \sqrt{116}}{8}\) and \(x = \frac{2 - \sqrt{116}}{8}\)

Key Concepts

Quadratic Equation

A quadratic equation is a special type of polynomial equation. Its most identifying feature is that the highest exponent of the variable is 2. This means every quadratic equation can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The reason \(a eq 0\) is that if \(a\) were zero, the equation would no longer be quadratic but linear instead.
Quadratic equations can arise in various real-life scenarios, such as calculating areas, determining projectile paths in physics, and finding maximum and minimum points in economic models. The crucial task when working with quadratic equations is often finding the solution or 'roots' of the equation, which represent the values of \(x\) that make the equation true.

Standard Form

The standard form of a quadratic equation is given by \(ax^2 + bx + c = 0\). This form is fundamental because it standardizes how quadratic equations are expressed and allows us to apply different algebraic techniques effectively.
To convert any quadratic equation into its standard form:

• Combine like terms if necessary.
• Move all terms to one side of the equation so that 0 is on the other side.

For example, in the equation \(4x^2 = 2x + 7\), we must first move all terms to the left to get \(4x^2 - 2x - 7 = 0\). This step is paramount as it prepares the equation for methods like factoring, completing the square, or using the quadratic formula.

Solving Quadratics

Solving quadratic equations involves finding the values of \(x\) that satisfy the equation. There are several methods to do this:

• Factoring: Where the quadratic can be expressed as a product of two binomials.
• Completing the Square: A procedure that alters the equation into a perfect square trinomial.
• Using the Quadratic Formula: A formula derived from completing the square, applicable to any quadratic equation.

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is a powerful tool because it can be used when other methods are impractical. In our exercise, rewriting the equation in standard form made it possible to identify \(a = 4\), \(b = -2\), and \(c = -7\), allowing straightforward application of the formula to find the solutions.

Algebraic Solutions

Applying the quadratic formula allows us to find algebraic solutions for \(x\). By plugging in the coefficients \(a\), \(b\), and \(c\) into the formula, we can solve for \(x\). In our case:\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 4 \times (-7)}}{2 \times 4}\]Simplifies to:\[x = \frac{2 \pm \sqrt{116}}{8}\]To further simplify, remember that \(\sqrt{116}\) can be broken down for easier computation, though exact value calculations often stay in square root form unless specified otherwise.
Finally, solving the simplified equation yields two potential solutions for \(x\):\[x = \frac{2 + \sqrt{116}}{8}\] and \[x = \frac{2 - \sqrt{116}}{8}\]These solutions provide the points where the quadratic equation crosses the x-axis on a graph, marking the critical solutions to our equation.