Problem 55 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ 5 x^{2}+1=6 x $$

Step-by-Step Solution

Answer

The solutions are \( x = 1 \) and \( x = 0.2 \).

Step 1: Move all terms to one side

Start by moving all terms to one side of the equation to set it to zero. Originally, the equation is \( 5x^2 + 1 = 6x \). Subtract \( 6x \) from both sides to obtain: \[ 5x^2 - 6x + 1 = 0 \] This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \) where \( a = 5 \), \( b = -6 \), and \( c = 1 \).

Step 2: Identify a, b, and c

Identify the coefficients and constant term from the quadratic equation \( 5x^2 - 6x + 1 = 0 \). Here, \( a = 5 \), \( b = -6 \), and \( c = 1 \).

Step 3: Apply the Quadratic Formula

Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), to find the solutions of the quadratic equation. Substitute \( a = 5 \), \( b = -6 \), and \( c = 1 \) into the formula: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 5 \times 1}}{2 \times 5} \].

Step 4: Simplify under the square root

Calculate the discriminant \( b^2 - 4ac \): \[ (-6)^2 - 4 \times 5 \times 1 = 36 - 20 = 16 \] This simplifies to \( \sqrt{16} = 4 \).

Step 5: Solve for x

Substitute \( \sqrt{16} = 4 \) back into the formula: \[ x = \frac{6 \pm 4}{10} \]. This splits into two solutions: \[ x_1 = \frac{6 + 4}{10} = 1 \] and \[ x_2 = \frac{6 - 4}{10} = 0.2 \].

Step 6: Conclusion

The solutions to the quadratic equation \( 5x^2 - 6x + 1 = 0 \) are \( x = 1 \) and \( x = 0.2 \).

Key Concepts

Quadratic Formula

Quadratic equations are in the form \( ax^2 + bx + c = 0 \). The magic ticket to solving these equations is the Quadratic Formula.This invaluable formula is expressed as:

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

The formula provides the solutions to any quadratic equation. By substituting the values of \( a \), \( b \), and \( c \), we can directly find the roots of the polynomial.
It’s crucial to recognize the coefficients correctly from the quadratic equation for the formula to work accurately. Identify \( a \), \( b \), and \( c \) from the standard form, equating it to zero first to ensure all terms are on one side.

• \( a \) is the coefficient of \( x^2 \).
• \( b \) is the coefficient of \( x \).
• \( c \) is the constant term.

Using these, you can calculate the solutions with ease. It simplifies the complex process, transforming it into straightforward substitution and arithmetic execution.

Discriminant

The discriminant is a special part of the quadratic formula. It determines the nature and number of roots a quadratic equation will have.The discriminant is given by:

• \( b^2 - 4ac \)

It's located under the square root in the quadratic formula.
The discriminant holds great importance, as it tells us:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is one real root (a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real roots. Instead, the roots are complex numbers.

For example, in the equation \( 5x^2 - 6x + 1 = 0 \), the discriminant calculation is \( 16 \), which indicates two distinct real roots.
Understanding the discriminant helps in predicting the solutions even before calculating them. It acts as a quick check for the type and number of solutions possible.

Roots of Polynomial

Finding the roots of a polynomial is essentially finding the values of \( x \) that satisfy the equation.For quadratic equations, these are often referred to as the solutions.
The roots represent the points where the parabola (the graph of a quadratic equation) intersects the x-axis.

• These intersections reveal the solutions of the quadratic equation \( ax^2 + bx + c = 0 \).

In our example equation, the roots are \( x = 1 \) and \( x = 0.2 \).
The interpretation of these roots can provide insights into the behavior of the function:

• When you graph the equation, the x-intercepts are at these specific root values.
• They determine where the function is equal to zero.

Hence, calculating the roots helps in understanding the nature of polynomial functions and their graphical behavior.