Question

Use factoring to solve the equation. Use a graphing calculator to check your solution if you wish. $$\frac{1}{5} x^{2}-2 x+5=0$$

Short answer

The solution for \( \frac{1}{5} x^{2} - 2 x + 5 = 0 \) is \( x = 5 \).

Step-by-step solution

Rewrite the equation in standard form

We write the given equation \( \frac{1}{5} x^{2} - 2 x + 5 = 0 \) by multiplying all terms by 5 to get the standard form of the equation which is \( x^{2} - 10 x + 25 = 0 \).

Factor the quadratic equation

To solve any quadratic equation, it can be factored into two binomial expressions if possible. The expression \( x^{2} - 10 x + 25 \) can be written as \( (x - 5)^{2} = 0 \).

Solve for the unknown

To find the values of x that satisfy the equation, set each factor equal to zero. So, \( x - 5 = 0 \) gives the solution \( x = 5 \).

Check the solution

Substitute \( x = 5 \) into the original equation \( \frac{1}{5} x^{2} - 2x + 5 \) and verify if the left-hand side equals the right-hand side. This step can also be done using a graphic calculator.

Key concepts

Standard Form of a Quadratic Equation

Understanding the standard form of a quadratic equation is crucial for solving and graphing them. The standard form is represented as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients, with \( a \) being non-zero. This form allows us to clearly identify the quadratic coefficient (\( a \) term), the linear coefficient (\( b \) term), and the constant term (\( c \) term).

For the given exercise, we begin by adjusting the equation into standard form: multiplying through by 5 transforms \( \frac{1}{5} x^2 - 2x + 5 \) into \( x^2 - 10x + 25 \) with no fractions involved. This makes factoring simpler and sets the stage for us to locate the roots of the equation easily.

Binomial Expressions

Binomial expressions are algebraic expressions containing two terms, such as \( x - 5 \). When a quadratic equation is factorable, we can express it as the product of two binomial expressions. Factoring involves finding two numbers that multiply to the constant term \( c \) and add to the linear coefficient \( b \) when \( a \) is 1, or are otherwise found through methods such as the AC method when \( a \) is not 1.

In our exercise, the equation \( x^2 - 10x + 25 \) is factorable into \( (x - 5)(x - 5) \) or \( (x - 5)^2 \) since \( -5 \) and \( -5 \) are the numbers that both add up to \( -10 \) and multiply to \( 25 \) which are the \( b \) and \( c \) coefficients, respectively.

Solving Quadratic Equations

To solve quadratic equations by factoring, we find values of the variable that make the equation equal to zero. After factoring, we set each binomial expression equal to zero and solve for the variable. This process gives us the roots or solutions of the equation.

In the context of our problem, the factored form is \( (x - 5)^2 = 0 \). We then set the binomial expression \( x - 5 \) equal to zero, leading to \( x = 5 \). This value of \( x \) is the solution to the equation, since substituting it back into the original equation yields \( 0 \) on both sides, confirming that \( x = 5 \) satisfies the equation.

Graphing Calculator Check

As a final step, it's a good practice to confirm our solutions using a graphing calculator. By entering the original equation, we can view its graph and visually validate the root(s). The x-intercepts of the graph correspond to our solutions.

For our example, plotting \( \frac{1}{5} x^2 - 2x + 5 \) would show us the parabola intersecting the x-axis at \( x = 5 \) if we solved correctly. This serves as a graphical double-check that reinforces our confidence in the solution. Moreover, in cases where equations are difficult to factor, graphing calculators are especially handy for obtaining approximate solutions through observation of the graph.