If λ is an integer and α, β are the roots of 4x 2– 16x + λ/4 = 0 such that 1

Question

If &lambda; is an integer and &alpha;, &beta; are the roots of 4x 2&ndash; 16x + &lambda;/4 = 0 such that 1 < &alpha; < 2 and 2 < &beta; < 3, then how many values can &lambda; take?

Prepp Team · Accepted Answer

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:

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

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.

Condition	Inequality	Result for $\lambda$
$f(1) > 0$	$-3 + \frac{\lambda}{16} > 0$	$\lambda > 48$
$f(2) < 0$	$-4 + \frac{\lambda}{16} < 0$	$\lambda < 64$
$f(3) > 0$	$-3 + \frac{\lambda}{16} > 0$	$\lambda > 48$

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

Concept	Description	Application in Problem
Quadratic Equation Roots	Values of $x$ that satisfy $ax^2+bx+c=0$.	Given roots $\alpha, \beta$ for $4x^2 – 16x + \frac{\lambda}{4} = 0$.
Parabola Direction	If $a>0$, parabola opens upwards. If $a<0$, downwards.	$x^2 – 4x + \frac{\lambda}{16} = 0$ has $a=1>0$, opens upwards.
Root Location Theory	Relates the position of roots relative to a point $k$ based on the sign of $f(k)$.	Used $f(1)>0$, $f(2)<0$, $f(3)>0$ based on $1<\alpha<2$ and $2<\beta<3$.
Solving Inequalities	Finding the range of values for a variable that satisfy an inequality.	Solved for $\lambda$ from $f(1)>0$, $f(2)<0$, $f(3)>0$.
Counting Integers in a Range	Number of integers between $a$ and $b$ (exclusive) is $b-a-1$. (Here it's inclusive of $a+1$ and $b-1$). Number of integers from $a$ to $b$ (inclusive) is $b-a+1$.	Counted integers from 49 to 63.

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.