Problem 112 - Algebra and Trigonometry

Question

If \(f(x)=5 x^{2}-2 x+9\) and \(f(a+1)=16,\) find the possible values for \(a\).

Step-by-Step Solution

Answer

The possible values of \(a\) are 0.4 and -2.

Step 1: Substitute the Value

Given the function \(f(x)=5x^2 - 2x + 9\) and knowing that \(f(a+1)=16\), substitute \(x = a+1\) into the function. Thus, \(f(a+1) = 5(a+1)^2 - 2(a+1) + 9\).

Step 2: Expand the Expression

Expand \(5(a+1)^2 - 2(a+1) + 9\). First, calculate \((a+1)^2\), which is \(a^2 + 2a + 1\). Then multiply by 5: \(5(a^2 + 2a + 1) = 5a^2 + 10a + 5\).

Step 3: Simplify the Equation

Now, subtract \(2(a+1)\), which gives \(2a + 2\). Combine all terms: \(5a^2 + 10a + 5 - 2a - 2 + 9\). Simplify it to \(5a^2 + 8a + 12\).

Step 4: Set Equation to 16

Since \(f(a+1) = 16\), set up the equation: \(5a^2 + 8a + 12 = 16\).

Step 5: Solve the Quadratic Equation

Subtract 16 from both sides to set the equation to zero: \(5a^2 + 8a + 12 - 16 = 0\), which simplifies to \(5a^2 + 8a - 4 = 0\).

Step 6: Use the Quadratic Formula

Use the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 5\), \(b = 8\), and \(c = -4\). Thus, solve for \(a\):

Step 7: Calculate the Discriminant

Calculate the discriminant: \(b^2 - 4ac = 8^2 - 4(5)(-4) = 64 + 80 = 144\).

Step 8: Find the Roots

The roots are \(a = \frac{-8 \pm \sqrt{144}}{10} = \frac{-8 \pm 12}{10}\). This gives two solutions: \(a = \frac{4}{10} = 0.4\) and \(a = \frac{-20}{10} = -2\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation used frequently in algebra. They are typically written in the form: \[ ax^2 + bx + c = 0 \]Here, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The quadratic equation forms a parabola when graphed, and solving it involves finding the values of \(x\) that satisfy the equation. Quadratic equations often appear in physics, engineering, and various real-world problems, making them very important to understand.

Function Substitution

When solving problems involving a function, you might need to substitute a variable. In this exercise, we are given a function \(f(x) = 5x^2 - 2x + 9\) and asked to find \(a\) such that \(f(a+1) = 16\). To do this, we substitute \(x\) with \(a+1\) in the function. This means we replace every occurrence of \(x\) in the equation with \(a+1\). For this example, \[ f(a+1) = 5(a+1)^2 - 2(a+1) + 9 \] Understanding function substitution is crucial as it allows us to transform and simplify equations, a common task in advanced math problems.

Discriminant Calculation

The discriminant is a key part of solving quadratic equations, found within the quadratic formula. It is represented by \(\Delta\) and is given by the expression: \[ b^2 - 4ac \] The discriminant tells us the nature of the roots of a quadratic equation:

• If \(\Delta > 0\), the quadratic equation has two distinct real roots.
• If \(\Delta = 0\), the quadratic equation has exactly one real root (a repeated root).
• If \(\Delta < 0\), the quadratic equation has two complex roots.

For the exercise in question, the values are \(a = 5\), \(b = 8\), and \(c = -4\). Calculating the discriminant: \[ 8^2 - 4 \times 5 \times (-4) = 64 + 80 = 144 \] With a discriminant of 144, there are two distinct real roots for the equation.

Quadratic Formula Application

The Quadratic Formula is a powerful tool for solving quadratic equations. It is used to find the values of \(x\) that satisfy the equation. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Using the quadratic formula involves several steps:

• First, identify the coefficients \(a\), \(b\), and \(c\) in your equation.
• Next, calculate the discriminant \(b^2 - 4ac\).
• Then, apply the quadratic formula using these values.

In our exercise, substituting these values into the quadratic formula gives us: \[a = \frac{-8 \pm \sqrt{144}}{10} = \frac{-8 \pm 12}{10}\] This results in two possible values for \(a\): \[ a = \frac{4}{10} = 0.4 \text{ and } a = \frac{-20}{10} = -2 \] Thus, the roots are 0.4 and -2. Using these steps, the quadratic formula allows for quick and reliable solutions to quadratic equations.