Problem 64 - Contemporary Precalculus

Question

$$\text {find the values of } x \text { for which } f(x)=g(x)$$. $$f(x)=3 x^{2}-x+5 ; \quad g(x)=x^{2}-2 x+26$$

Step-by-Step Solution

Answer

Question: For the functions f(x) = 3x^2 - x + 5 and g(x) = x^2 - 2x + 26, determine the values of x for which f(x) = g(x). Answer: The values of x for which f(x) = g(x) are x = 3 and x = -3.5.

Step 1: Set f(x) equal to g(x)

To find the values of x for which f(x) = g(x), we set f(x) and g(x) equal to each other: $$3x^2 - x + 5 = x^2 - 2x + 26$$

Step 2: Move all terms to one side of the equation

Next, we move all the terms from the right side to the left side of the equation by subtracting both sides by \(x^2 - 2x + 26\): $$3x^2 - x + 5 - (x^2 - 2x + 26) = 0$$

Step 3: Simplify the equation

Now, we simplify the expression on the left side of the equation: $$3x^2 - x + 5 - x^2 + 2x - 26 = 0$$ $$(3x^2 - x^2) + (-x + 2x) + (5 - 26) = 0$$ $$2x^2 + x - 21 = 0$$

Step 4: Solve the quadratic equation

We are left with a quadratic equation, which we need to solve for x. We can use either factoring (if possible), completing the square, or the quadratic formula. In this case, we will be using the quadratic formula, which is given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For our equation \(2x^2 + x - 21 = 0\), we have coefficients a = 2, b = 1, and c = -21. Plugging these values into the quadratic formula, we get: $$x = \frac{-(1) \pm \sqrt{(1)^2 - 4(2)(-21)}}{2(2)}$$

Step 5: Calculate the discriminant

Next, we calculate the discriminant, which is the part under the square root in the quadratic formula: $$\Delta = b^2 - 4ac = (1)^2 - 4(2)(-21) = 1 + 168 = 169$$

Step 6: Solve for x

Now that we have the discriminant and the quadratic formula set up, we can solve for x: $$x = \frac{-(1) \pm \sqrt{169}}{4}$$ $$x = \frac{-1 \pm 13}{4}$$ Now we have two possible solutions for x: $$x = \frac{-1 + 13}{4} = \frac{12}{4} = 3$$ $$x = \frac{-1 - 13}{4} = \frac{-14}{4} = -3.5$$ The values of x for which f(x) = g(x) are x = 3 and x = -3.5.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. Quadratic equations are polynomials of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The quadratic formula allows us to find the roots of these equations. It is given by:

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

The formula uses three coefficients from the quadratic equation:

• \( a \): coefficient of \( x^2 \)
• \( b \): coefficient of \( x \)
• \( c \): constant term

This method is particularly useful when the equation cannot be easily factored. By substituting the coefficients from the polynomial, the formula provides a straightforward path to the solution.

Discriminant

The discriminant is a key component of the quadratic formula, found under the square root: \( b^2 - 4ac \). It tells us about the nature of the roots of the quadratic equation:

• If the discriminant is positive, \( b^2 - 4ac > 0 \): There are two distinct real roots.
• If the discriminant is zero, \( b^2 - 4ac = 0 \): There is exactly one real root, also known as a double root.
• If the discriminant is negative, \( b^2 - 4ac < 0 \): There are no real roots, but two complex roots.

In our equation, the discriminant is calculated as \( 169 \), a positive number, indicating two distinct real solutions.

Polynomial Equations

Polynomial equations are expressions that involve sums of powers of a variable, like \( x \). A quadratic equation is a specific type of polynomial equation of degree 2, such as \( ax^2 + bx + c = 0 \). Solving these equations often involves different methods:

• Factoring: Finding factors (if they exist) that, when multiplied, give the original polynomial.
• Completing the square: Turning the equation into a perfect square trinomial.
• Quadratic formula: Direct calculation using \( a \), \( b \), and \( c \) from the equation.

Understanding the form and degree of these equations is essential for determining the best approach to solve them.

Roots of Equations

Roots of equations are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). They are the points where the polynomial crosses the x-axis. For quadratic equations, the roots can be real or complex, as determined by the discriminant:

• Real roots: When the solution involves real numbers.
• Complex roots: When the solution involves imaginary numbers, resulting from a negative discriminant.

In our case, the equation \( 2x^2 + x - 21 = 0 \) has roots at \( x = 3 \) and \( x = -3.5 \), which means these are the values that satisfy \( f(x) = g(x) \) for the given functions.