Question

\(4 z^2+28 z=15\)

Short answer

The solutions for \(z\) are approximately -0.5364 and -7.0

Step-by-step solution

Rearrange in the form of \(ax^2 + bx + c = 0\)

Rearrange the equation \(4 z^2+28 z = 15\) to the form \(ax^2+bx+c=0\), we get \(4 z^2+28 z - 15 = 0\)

Apply the Quadratic Formula

Apply the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=4\), \(b=28\), \(c=-15\) to solve for \(z\)

Compute the Solution

Plug \(a=4\), \(b=28\), \(c=-15\) into the quadratic formula gives us two solutions: \(z1 = \frac{-28 + \sqrt{28^2 - 4*4*(-15)}}{2*4} = -0.5364\) and \(z2 = \frac{-28 - \sqrt{28^2 - 4*4*(-15)}}{2*4} = -7.0\). Please note that the results may slightly vary due to rounding error.

Key concepts

Quadratic Formula

The quadratic formula is a fundamental tool for solving quadratic equations. This formula allows us to find the roots (or solutions) of any quadratic equation, which is in the standard form of \( ax^2 + bx + c = 0 \). The quadratic formula is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation. The formula works by calculating the discriminant, \( b^2 - 4ac \), which tells us about the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root.
• If it is negative, the roots are complex numbers.

By using the quadratic formula, we can easily solve quadratic equations even when factors are complex or non-integers. Remember, careful computation of the discriminant is crucial!

Polynomial Equations

Polynomial equations are equations that involve variables raised to various powers. The equation in question, \(4z^2 + 28z = 15\), is a polynomial of the second degree because the highest power of the variable (\(z\)) is 2. Polynomial equations can have several forms depending on the degree of the polynomial:

• Linear equations have the highest power of 1.
• Quadratic equations have the highest power of 2.
• Cubic equations have the highest power of 3, and so forth.

The structure and complexity of polynomial equations increase with the degree. Quadratic equations, like the one we're discussing, are foundational in algebra because they frequently appear across different fields such as physics, engineering, and economics. Their solutions can exhibit a parabola when graphed, depending on the coefficients \(a\), \(b\), and \(c\). Solving polynomial equations often requires rearranging them to a standard form so tools like the quadratic formula can be used effectively.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of the variable that satisfy the equation. Several methods exist:

• Factoring
• Using the quadratic formula
• Completing the square

In this scenario, we use the quadratic formula, benefiting from its versatility and ability to handle any quadratic equation regardless of its complexity. After rearranging the given polynomial \( 4z^2 + 28z = 15 \) into its standard form, we plug the coefficients into the quadratic formula.The steps involve computing the discriminant, \( b^2 - 4ac \), which dictates the roots' nature. After evaluating the discriminant, you calculate the roots using the plus-minus function to find both potential solutions. Each solution corresponds to a different potential intersection point on the parabola represented by the quadratic equation. This approach ensures that you capture all possible solutions, even when factoring techniques may fall short, particularly with non-integer or complex roots.

Algebraic Expressions

Algebraic expressions form the backbone of algebra, consisting of variables and constants combined with operations such as addition, subtraction, multiplication, and division. In our problem, the expression \(4z^2 + 28z - 15 = 0\) is an algebraic expression set equal to zero. Key elements of algebraic expressions include:

• Terms: Parts of the expression separated by plus or minus signs. For instance, in \(4z^2 + 28z - 15\), there are three terms.
• Coefficients: Numbers multiplying the variables (e.g., 4 in \(4z^2\)).
• Constants: Numbers without variables (e.g., -15).

Solving an equation involves simplifying and manipulating these terms to find the variable's values that satisfy the original equation. Algebraic proficiency allows one to rearrange equations, recognize patterns, and apply formulas accurately to obtain solutions, as demonstrated in this problem using the quadratic formula. Understanding these expressions also simplifies the solution process and enhances one's ability to tackle more complex algebraic problems.