Problem 81 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ 3 y^{2}-14 y-5=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = 5 \) and \( y = -\frac{1}{3} \).

Step 1: Identify the type of equation

The given equation is a quadratic equation in the form \(ax^2 + bx + c = 0\). Here, \(a = 3\), \(b = -14\), and \(c = -5\).

Step 2: Apply the quadratic formula

The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We will use this formula to solve for \(y\).

Step 3: Calculate the discriminant

The discriminant is given by \( b^2 - 4ac \). Substitute the values: \((-14)^2 - 4 \cdot 3 \cdot (-5) = 196 + 60 = 256\). The discriminant is 256.

Step 4: Evaluate the expression under the square root

Since the discriminant is 256, calculate \( \sqrt{256} = 16 \).

Step 5: Solve for the two possible values of y

Substitute \( b = -14 \), \( \sqrt{256} = 16 \), and \( a = 3 \) into the quadratic formula: 1. \( y = \frac{-(-14) + 16}{2 \times 3} = \frac{14 + 16}{6} = \frac{30}{6} = 5 \) 2. \( y = \frac{-(-14) - 16}{2 \times 3} = \frac{14 - 16}{6} = \frac{-2}{6} = -\frac{1}{3} \)

Step 6: State the solutions

The solutions to the equation are \( y = 5 \) and \( y = -\frac{1}{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations. A quadratic equation typically takes the form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). To find the solutions for \( x \), we employ the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula enables us to find the roots of the equation by allowing us to calculate both possible solutions efficiently.

• "\(-b\)" becomes the opposite of \( b \).
• "\(\pm\)" indicates there are usually two solutions: one where you add the square root, and another where you subtract.
• The value under the square root - "\(b^2 - 4ac\)" - is called the discriminant.

Plug in the values from your specific equation into \( a \), \( b \), and \( c \) and solve for \( x \) using this formula to find your solution.

Discriminant

The discriminant is a component of the quadratic formula represented by the expression \( b^2 - 4ac \). It plays a significant role in determining the nature of the roots of a quadratic equation. Here's what the discriminant can tell us:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions. This indicates that the parabola crosses the x-axis at two points.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution. The parabola touches the x-axis at exactly one point, resulting in a "double root".
• If \( b^2 - 4ac < 0 \), there are no real solutions, meaning the parabola does not intersect the x-axis.

In our example, with values \( b = -14 \), \( a = 3 \), and \( c = -5 \), the discriminant was calculated to be \( 256 \). Since \( 256 > 0 \), this confirms the existence of two distinct real solutions for the equation.

Solving Quadratic Equations

Solving quadratic equations becomes much easier by understanding and applying both the quadratic formula and the discriminant. Here's a quick guide to solving these equations:1. **Identify the equation type.**
In a typical scenario, you have \( ax^2 + bx + c = 0 \). Recognize your constants \( a \), \( b \), and \( c \) from your equation. 2. **Calculate the discriminant.**
Use \( b^2 - 4ac \) to assess the roots' nature and number. This step precludes deeper calculation if no real roots exist. 3. **Apply the quadratic formula.**
Insert \( a \), \( b \), and \( c \) into: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Compute the values for \( x \) for both \( +\sqrt{} \) and \( -\sqrt{} \) scenarios to find the two solutions.In our example, when we calculated step-by-step for \( y \), we obtained the solutions \( y = 5 \) and \( y = -\frac{1}{3} \). The use of these steps can systematically solve similar quadratic equations you face in your studies.