Problem 62 - Contemporary Precalculus

Question

$$\text {find the values of } x \text { for which } f(x)=g(x)$$. $$f(x)=2 x^{2}+4 x-4 ; \quad g(x)=x^{2}+12 x+6$$

Step-by-Step Solution

Answer

Question: Determine the values of \(x\) for which the functions \(f(x) = 2x^2 + 4x - 4\) and \(g(x) = x^2 + 12x + 6\) are equal. Answer: The values of \(x\) for which \(f(x) = g(x)\) are \(x = \frac{8+\sqrt{104}}{2}\) and \(x = \frac{8-\sqrt{104}}{2}\).

Step 1: Combine like terms

First, we need to bring all terms to one side to form a quadratic equation. We do this by subtracting \(x^2\) and \(12x\) from both sides and adding 6 to both sides, getting: $$x^2 - 8x - 10 = 0$$

Step 2: Determine the method to solve the quadratic

Now, we need to determine how to solve this quadratic equation. It doesn't look like this quadratic can be easily factored, so we can use the quadratic formula instead: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In our case, \(a = 1, b = -8,\) and \(c = -10\).

Step 3: Solve the quadratic equation for \(x\)

Using the quadratic formula with the coefficients from our equation, we get: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-10)}}{2(1)}$$ Simplifying the expression under the square root, we get: $$x = \frac{8 \pm \sqrt{64 + 40}}{2}$$ $$x = \frac{8 \pm \sqrt{104}}{2}$$ So the solutions for \(x\) are: $$x = \frac{8+\sqrt{104}}{2} \quad \text{and} \quad x = \frac{8-\sqrt{104}}{2}$$ These are the values for \(x\) for which \(f(x) = g(x)\).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool in solving quadratic equations, especially when factoring is not straightforward. It provides a solution to any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula itself is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The quadratic formula uses three coefficients: \(a\), \(b\), and \(c\). These are simply the numerical factors in front of \(x^2\), \(x\), and the constant term in the equation, respectively. Here’s a step-by-step guide to using the quadratic formula:

• Identify \(a\), \(b\), and \(c\) from the quadratic equation.
• Plug these values into the formula.
• Calculate the discriminant, \(b^2 - 4ac\).
• Use the discriminant to find the value(s) of \(x\).

Finding the square root of the discriminant is critical. If it's positive, you get two real solutions, if zero, one real solution, and if negative, no real solutions. This versatility makes the quadratic formula a powerful method.

Factoring Quadratics

Factoring quadratics is another method for finding the roots of quadratic equations. While not always applicable (like in this exercise where factoring was not easy), it is quite effective when the quadratic can be easily expressed as a product of two binomials.To factor a quadratic equation of the form \(ax^2 + bx + c = 0\):

• Look for two numbers that multiply to \(ac\) and add up to \(b\).
• Use these numbers to split the middle term, \(bx\), then factor by grouping.
• Check your factor pairs to ensure correctness.

Factoring is often the preferred method if it’s straightforward because it’s quick and doesn’t involve complex calculations, unlike the quadratic formula. However, if the quadratic cannot be factored easily into integers, then other methods like the quadratic formula or completing the square are used instead.

Solving Equations

Solving quadratic equations involves methods for finding \(x\) such that the equation holds true. For the equation given in the problem, \(f(x) = g(x)\), the ultimate goal is to find values of \(x\) that make this equality true.In the process seen in the exercise, we first combined like terms from both functions into a single quadratic equation. This is a fundamental step because solving a simpler form (\(x^2 - 8x - 10 = 0\)) is much easier. Here’s a general approach:

• Rearrange the equation so all terms are on one side and zero is on the other.
• Decide on the method of solving: factoring, quadratic formula, or other means.
• Apply the method chosen to find the solution for \(x\).

Checking solutions by substituting them back into the original functions ensures accuracy. Consistency in solution validation is important because mistakes in calculations can lead to errors.