Understanding the Quadratic Equation and Root Intervals

The given question involves a quadratic equation and conditions on its roots. We are given the equation \(4x^2 - 16x + \frac{\lambda}{4} = 0\), where \(\lambda\) is an integer. The roots of this equation, denoted by \(\alpha\) and \(\beta\), satisfy the conditions \(1 < \alpha < 2\) and \(2 < \beta < 3\). We need to determine how many possible integer values \(\lambda\) can take under these conditions.

First, let's simplify the quadratic equation by dividing by 4:

\(\frac{4x^2 - 16x + \frac{\lambda}{4}}{4} = \frac{0}{4}\)

\(x^2 - 4x + \frac{\lambda}{16} = 0\)

Let \(f(x) = x^2 - 4x + \frac{\lambda}{16}\). The roots of \(f(x)=0\) are \(\alpha\) and \(\beta\). Since the coefficient of \(x^2\) is positive (which is 1), the parabola representing \(f(x)\) opens upwards.

Applying Conditions on Root Location

The conditions \(1 < \alpha < 2\) and \(2 < \beta < 3\) tell us that one root lies strictly between 1 and 2, and the other root lies strictly between 2 and 3. This means that \(x=2\) lies between the two roots \(\alpha\) and \(\beta\).

For a quadratic function \(f(x) = ax^2 + bx + c\) with \(a > 0\), if a value \(k\) is between the roots, then \(f(k)\) must be negative. If a value \(k\) is outside the roots, \(f(k)\) must be positive.

In our case, \(a=1 > 0\). The roots are \(\alpha\) and \(\beta\). The conditions are \(1 < \alpha < 2\) and \(2 < \beta < 3\). This implies:

• The value \(x=1\) is less than the smaller root \(\alpha\). Thus, \(f(1)\) must be positive.
• The value \(x=2\) is between the roots \(\alpha\) and \(\beta\). Thus, \(f(2)\) must be negative.
• The value \(x=3\) is greater than the larger root \(\beta\). Thus, \(f(3)\) must be positive.

Let's calculate \(f(1)\), \(f(2)\), and \(f(3)\) using the function \(f(x) = x^2 - 4x + \frac{\lambda}{16}\):

• \(f(1) = (1)^2 - 4(1) + \frac{\lambda}{16} = 1 - 4 + \frac{\lambda}{16} = -3 + \frac{\lambda}{16}\)
• \(f(2) = (2)^2 - 4(2) + \frac{\lambda}{16} = 4 - 8 + \frac{\lambda}{16} = -4 + \frac{\lambda}{16}\)
• \(f(3) = (3)^2 - 4(3) + \frac{\lambda}{16} = 9 - 12 + \frac{\lambda}{16} = -3 + \frac{\lambda}{16}\)

Setting up Inequalities for Lambda

Now we use the conditions derived from the root locations:

1. \(f(1) > 0 \implies -3 + \frac{\lambda}{16} > 0\)
2. \(f(2) < 0 \implies -4 + \frac{\lambda}{16} < 0\)
3. \(f(3) > 0 \implies -3 + \frac{\lambda}{16} > 0\)

Let's solve each inequality for \(\lambda\):

From inequality 1:

\(-3 + \frac{\lambda}{16} > 0\)

\(\frac{\lambda}{16} > 3\)

\(\lambda > 3 \times 16\)

\(\lambda > 48\)

From inequality 2:

\(-4 + \frac{\lambda}{16} < 0\)

\(\frac{\lambda}{16} < 4\)

\(\lambda < 4 \times 16\)

\(\lambda < 64\)

Inequality 3 is the same as inequality 1, so it also gives \(\lambda > 48\).

Finding the Range and Counting Integer Values of Lambda

Combining the inequalities \(\lambda > 48\) and \(\lambda < 64\), we get the range for \(\lambda\) as \(48 < \lambda < 64\).

We are given that \(\lambda\) is an integer. The integers strictly between 48 and 64 are \(49, 50, 51, \dots, 63\).

To find the number of integers in this range, we can subtract the smallest integer from the largest integer and add 1:

Number of values = \(63 - 49 + 1 = 14 + 1 = 15\)

Therefore, \(\lambda\) can take 15 different integer values.

The combined condition is \(48 < \lambda < 64\). Since \(\lambda\) must be an integer, the possible values are \(49, 50, \dots, 63\). There are 15 such integer values.

Revision Table: Key Concepts

Additional Information: Properties of Quadratic Functions

Understanding the graph of a quadratic function \(f(x) = ax^2 + bx + c\) is crucial for solving problems related to root location. The roots are the x-intercepts of the graph.

• If \(a > 0\), the parabola has a minimum point (vertex). The function is negative between the roots and positive outside the roots.
• If \(a < 0\), the parabola has a maximum point (vertex). The function is positive between the roots and negative outside the roots.
• The vertex of the parabola is at \(x = -\frac{b}{2a}\). For \(f(x) = x^2 - 4x + \frac{\lambda}{16}\), the vertex is at \(x = -\frac{(-4)}{2(1)} = 2\). Notice that the intervals \((1, 2)\) and \((2, 3)\) are centered around the vertex location, which is why \(f(1) = f(3)\).
• The Discriminant (\(\Delta = b^2 - 4ac\)) tells us about the nature of the roots:
  • \(\Delta > 0\): Two distinct real roots.
  • \(\Delta = 0\): Exactly one real root (a repeated root).
  • \(\Delta < 0\): No real roots (two complex conjugate roots).
  In this problem, since we have two distinct real roots \(\alpha\) and \(\beta\), the discriminant must be positive. For \(x^2 - 4x + \frac{\lambda}{16} = 0\), \(\Delta = (-4)^2 - 4(1)(\frac{\lambda}{16}) = 16 - \frac{\lambda}{4}\). For distinct real roots, \(16 - \frac{\lambda}{4} > 0 \implies 16 > \frac{\lambda}{4} \implies 64 > \lambda\). Our derived condition \(48 < \lambda < 64\) satisfies this requirement.