Problem 8 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 4 z^{2}=7-27 z $$

Step-by-Step Solution

Answer

Solutions are \(z = \frac{1}{4}\) (with check error) and \(z = -7\).

Step 1: Write the equation in standard form

To solve the quadratic equation \(4z^2 = 7 - 27z\), we first need to write it in standard form. This means moving all terms to one side so that the equation becomes zero on the other side. Subtract \(7\) and add \(27z\) to both sides of the equation to get: \(4z^2 + 27z - 7 = 0\).

Step 2: Identify coefficients

The standard form of a quadratic equation is \(az^2 + bz + c = 0\). From the equation \(4z^2 + 27z - 7 = 0\), we can identify the coefficients: \(a = 4\), \(b = 27\), and \(c = -7\).

Step 3: Use the quadratic formula

To solve using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), substitute the values of \(a\), \(b\), and \(c\) into the formula: \[ z = \frac{-27 \pm \sqrt{27^2 - 4(4)(-7)}}{2(4)} \].

Step 4: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\): \[ 27^2 - 4 \cdot 4 \cdot (-7) = 729 + 112 = 841 \]. Since 841 is a perfect square, the roots are real and rational.

Step 5: Simplify the expression

Evaluate the square root of the discriminant and substitute back: \(\sqrt{841} = 29\). The equation is now: \[ z = \frac{-27 \pm 29}{8} \].

Step 6: Find the two possible values for z

Solve for \(z\) by considering \(-27 + 29\) and \(-27 - 29\):1. \(z = \frac{-27 + 29}{8} = \frac{2}{8} = \frac{1}{4}\)2. \(z = \frac{-27 - 29}{8} = \frac{-56}{8} = -7\).

Step 7: Verify the solutions

Substitute each solution back into the original equation to verify.For \(z = \frac{1}{4}\): \(4(\frac{1}{4})^2 + 27(\frac{1}{4}) - 7 = 1 + 6.75 - 7 = 0.75 eq 0\) (error indicates a need to re-check arithmetic or steps in verification).For \(z = -7\):\(4(-7)^2 + 27(-7) - 7 = 196 - 189 - 7 = 0\) confirming it works.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions, also known as roots, of a quadratic equation. A quadratic equation is typically of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Using this formula allows you to determine the values of \(x\) that satisfy the quadratic equation. It is an essential technique because it works for any quadratic equation, regardless of its type. Here's a brief breakdown of how it works:

• The symbol \(\pm\) indicates that there are generally two solutions (one for the \(+\) and one for the \(-\) operation). This is because quadratics may have up to two real roots.
• The expression under the square root sign, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots.

The quadratic formula simplifies the process of solving quadratic equations, as long as you correctly plug in the values for \(a\), \(b\), and \(c\) from the standard form of the equation.

Discriminant

The discriminant is a key part of the quadratic formula and plays a vital role in determining the nature of the roots of the quadratic equation.
It is the part of the formula under the square root: \(b^2 - 4ac\). Depending on the value of the discriminant, the roots of the quadratic equation will vary:

• If the discriminant is positive, \(b^2 - 4ac > 0\), the equation has two distinct real roots. This case occurs when the value under the square root is greater than zero, resulting in two different solutions for \(x\).
• If the discriminant is zero, \(b^2 - 4ac = 0\), the equation has exactly one real root. This means both solutions are the same, providing a single repeated root for \(x\).
• If the discriminant is negative, \(b^2 - 4ac < 0\), the equation has no real roots; instead, the roots are complex or imaginary. This occurs when the square root of a negative number needs to be calculated, leading to complex solutions.

Understanding the discriminant helps predict what kind of solutions a quadratic equation has without fully solving it.

Standard Form of Quadratic Equations

The standard form of a quadratic equation is crucial for applying the quadratic formula and understanding the equation's structure.
A quadratic equation is in standard form when written as:
\[ ax^2 + bx + c = 0 \]
In this form:

• \(a\) is the coefficient of \(x^2\), and it is crucial that \(a eq 0\) because, without it, the equation wouldn't be quadratic.
• \(b\) is the coefficient of \(x\).
• \(c\) is the constant term, which does not involve \(x\).

When given a quadratic equation, the first task is often to rearrange it into this standard form by moving all the terms to one side of the equation, resulting in zero on the other. This makes it clear what the values of \(a\), \(b\), and \(c\) are, which are needed for using the quadratic formula.
For example, if you begin with an equation like \(4z^2 = 7 - 27z\), bringing all terms to one side gives \(4z^2 + 27z - 7 = 0\), making it easy to identify that \(a = 4\), \(b = 27\), and \(c = -7\). Getting the equation into standard form is integral because it is the starting point for solving quadratic equations.