Answered

A quadratic function is defined as $f(x)=2 x^2-3 x-5$. For what positive integer value of $x$ is $f(x)$ equal to 15?

Verified Answer :

  • Rewrite the given function equality as the quadratic equation \(2x^2 - 3x - 20 = 0\).
  • Apply the quadratic formula to solve for \(x\), yielding roots \(4\) and \(-2.5\).
  • Evaluate which root meets the condition of being a positive integer.
  • Conclude that the positive integer solution is \(x = \boxed{4}\).

Explanation

  1. Problem Analysis

The function is defined as \(f(x) = 2x^2 - 3x - 5\). We are asked to find a positive integer value of \(x\) such that the value of the function equals 15, which means we must solve the equation \(f(x) = 15\). Substituting the given expression for \(f(x)\), we have:
\[2x^2 - 3x - 5 = 15\]

  1. Rearranging the Quadratic Equation

To solve the equation \(2x^2 - 3x - 5 = 15\), we first rearrange it into the standard quadratic form \(ax^2 + bx + c = 0\). By subtracting 15 from both sides, we get:
\[2x^2 - 3x - 5 - 15 = 0\]
\[2x^2 - 3x - 20 = 0\]

  1. Solving the Quadratic Equation

We can solve the quadratic equation \(2x^2 - 3x - 20 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = -3\), and \(c = -20\). Substituting these values:
\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-20)}}{2(2)}\]
\[x = \frac{3 \pm \sqrt{9 + 160}}{4}\]
\[x = \frac{3 \pm \sqrt{169}}{4}\]
\[x = \frac{3 \pm 13}{4}\]

  1. Identifying the Correct Solution

Calculating the two possible values for \(x\):

  1. \(x_1 = \frac{3 + 13}{4} = \frac{16}{4} = 4\)
  2. \(x_2 = \frac{3 - 13}{4} = \frac{-10}{4} = -2.5\)

Since the problem requires a positive integer value for \(x\), we check both results. The value \(x = 4\) is a positive integer, whereas \(x = -2.5\) is not. Therefore, the only valid solution is \(x = 4\).

Examples

Quadratic equations frequently appear in modeling physical phenomena, such as the path of a projectile. If a person throws a ball into the air, its height \(h\) at time \(t\) might be described by a quadratic function. Finding a specific height at a given time is a common problem, similar to finding the value of \(x\) for a specific function output. Understanding how to solve quadratic equations allows one to predict these events accurately and solve problems in engineering, physics, and economics.

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