Problem 43 - Algebra 1: Concepts and Skills

Question

Use the quadratic formula to solve the equation. If the solution involves radicals, round to the nearest hundredth. $$-3 y^{2}+2 y+8=0$$

Step-by-Step Solution

Answer

The solutions for the given equation are \(y = -1.33\) and \(y = 2\).

Step 1: Identify coefficients

In the given quadratic equation, the coefficients correlate to the equation's format \(ax^2 + bx + c = 0\), where \(a = -3\), \(b = 2\), and \(c = 8\).

Step 2: Insert coefficients into the quadratic formula

Substitute \(a, b, c\) with real values into the quadratic formula. It will be: \(y = {-2 \pm \sqrt{2^2 - 4*(-3)*8} \over 2*(-3)}\).

Step 3: Simplify inside the square root

Perform the operation inside the square root first, according to the BIDMAS rule (Brackets, Indices, Division and Multiplication, and finally, Addition and Subtraction). The equation becomes: \(y = {-2 \pm \sqrt{4+96} \over -6}\).

Step 4: Further Simplification

Simplify the square root. The equation becomes: \(y = {-2 \pm \sqrt{100} \over -6}\), which simplifies to \(y = {-2 \pm 10 \over -6}\).

Step 5: Obtain the solutions

This equation gives us two solutions; one with the plus sign and the other with the minus sign. So, the solutions are: \(y = {-2 + 10 \over -6} = -1.33\) (rounded to the nearest hundredth), and \(y = {-2 - 10 \over -6} = 2\) (rounded to the nearest hundredth).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra. They take the form \( ax^2 + bx + c = 0 \). In this formula, \(x\) represents the variable or unknown, while \(a\), \(b\), and \(c\) are coefficients with \(a eq 0\). The inclusion of the square term \(ax^2\) is what makes these equations 'quadratic'.

Quadratic equations are prevalent in various fields, including physics, engineering, and economics, because they can describe parabolic paths, such as trajectory or graphing profit and loss. When solving these equations, the goal is to find the value(s) of \(x\) that make the equation true.

One of the most powerful methods to solve these equations is using the Quadratic Formula. This formula acts like a solver for any quadratic equation, given the quadratic is not completely degenerate. Understanding the structure of the equation helps to correctly identify and apply these coefficients into the formula.

Roots of Equations

The roots or solutions of a quadratic equation are the values that fulfill the equation \( ax^2 + bx + c = 0 \). The term 'roots' comes from the idea that these are the values that "ground" or resolve the equation fully.

In a quadratic equation, there can be one, two, or no real roots. This depends on the discriminant, \( b^2 - 4ac \). When this value is:

• Positive: There are two distinct real roots.
• Zero: There is exactly one real root (also called a repeated or double root).
• Negative: There are no real roots, but instead two complex roots.

In the exercise given, the discriminant is 100, which is positive, indicating two real roots. By calculating these roots, we can find the specific values of the variable \(y\) that satisfy the equation fully.

Algebraic Solutions

Solving quadratic equations algebraically is a critical skill in mathematics. For the problem at hand, we use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This provides a straightforward way to find the roots, aligned with the coefficients identified from the equation.

Firstly, identifying the correct coefficients \(a = -3\), \(b = 2\), \(c = 8\) was imperative. Substituting these into the quadratic formula allows us to calculate the roots. Consideration of the signs and careful arithmetic is crucial.

In calculations, simplifying the expression under the square root and then the overall fraction is essential. After simplifying, we get two potential solutions, derived from the 'plus/minus' (\(\pm\)) aspect of the equation. This clearly provides both potential real solutions: \(y = -1.33\) and \(y = 2\), as seen in the problem's step-by-step solution.