If both the roots of the equation x 2– 2kx + k 2- 4 = 0 lie between – 3 and 5, then which one of the following is correct?
– 1 < k < 3
Understanding the Problem: Roots of a Quadratic Equation
The question asks us to find the range of values for the parameter $k$ such that both roots of the given quadratic equation $x^2 - 2kx + k^2 - 4 = 0$ lie strictly between $-3$ and $5$.
Let the quadratic function be $f(x) = x^2 - 2kx + k^2 - 4$. For the roots of a quadratic equation $ax^2 + bx + c = 0$ to lie between two specific numbers $p$ and $q$ ($p < q$), several conditions must be satisfied. Here, $a=1$, $b=-2k$, $c=k^2-4$, $p=-3$, and $q=5$.
Conditions for Roots Between Two Values
For the roots of $ax^2 + bx + c = 0$ to lie between $p$ and $q$ ($p < q$), the following conditions must hold:
- The discriminant $D$ must be non-negative, i.e., $D \ge 0$, for real roots.
- $a \cdot f(p) > 0$
- $a \cdot f(q) > 0$
- The axis of symmetry, $\frac{-b}{2a}$, must lie between $p$ and $q$, i.e., $p < \frac{-b}{2a} < q$.
Applying the Conditions to the Given Equation
Condition 1: Discriminant $D \ge 0$
The discriminant is $D = b^2 - 4ac$.
\[D = (-2k)^2 - 4(1)(k^2 - 4)\] \[D = 4k^2 - 4k^2 + 16\] \[D = 16\]Since $D = 16$, which is strictly greater than 0, the equation always has two distinct real roots. This condition is always satisfied for any real value of $k$.
Condition 2: $a \cdot f(-3) > 0$
Here $a=1$, so we need $f(-3) > 0$.
\[f(x) = x^2 - 2kx + k^2 - 4\] \[f(-3) = (-3)^2 - 2k(-3) + k^2 - 4\] \[f(-3) = 9 + 6k + k^2 - 4\] \[f(-3) = k^2 + 6k + 5\]We need $k^2 + 6k + 5 > 0$.
Factor the quadratic expression:
\[(k+1)(k+5) > 0\]This inequality holds when both factors are positive or both are negative. This occurs when $k < -5$ or $k > -1$.
So, the condition $f(-3) > 0$ gives $k \in (-\infty, -5) \cup (-1, \infty)$.
Condition 3: $a \cdot f(5) > 0$
Here $a=1$, so we need $f(5) > 0$.
\[f(5) = (5)^2 - 2k(5) + k^2 - 4\] \[f(5) = 25 - 10k + k^2 - 4\] \[f(5) = k^2 - 10k + 21\]We need $k^2 - 10k + 21 > 0$.
Factor the quadratic expression:
\[(k-3)(k-7) > 0\]This inequality holds when both factors are positive or both are negative. This occurs when $k < 3$ or $k > 7$.
So, the condition $f(5) > 0$ gives $k \in (-\infty, 3) \cup (7, \infty)$.
Condition 4: Axis of Symmetry Between -3 and 5
The axis of symmetry is given by $\frac{-b}{2a}$.
\[\text{Axis of symmetry} = \frac{-(-2k)}{2(1)} = \frac{2k}{2} = k\]We need the axis of symmetry to be between $-3$ and $5$:
\[-3 < k < 5\]So, this condition gives $k \in (-3, 5)$.
Finding the Intersection of all Conditions
To satisfy all conditions simultaneously, $k$ must be in the intersection of the intervals obtained from each condition:
- From $D \ge 0$: $k \in \mathbb{R}$
- From $f(-3) > 0$: $k \in (-\infty, -5) \cup (-1, \infty)$
- From $f(5) > 0$: $k \in (-\infty, 3) \cup (7, \infty)$
- From $-3 < k < 5$: $k \in (-3, 5)$
Let's find the intersection step by step.
Intersection of conditions 2 and 4:
\[((-\infty, -5) \cup (-1, \infty)) \cap (-3, 5)\] \[= ((-\infty, -5) \cap (-3, 5)) \cup ((-1, \infty) \cap (-3, 5))\] \[= \emptyset \cup (-1, 5)\] \[= (-1, 5)\]So, from conditions 2 and 4, we get $k \in (-1, 5)$.
Now, intersect this result with condition 3:
\[(-1, 5) \cap ((-\infty, 3) \cup (7, \infty))\] \[= ((-1, 5) \cap (-\infty, 3)) \cup ((-1, 5) \cap (7, \infty))\] \[= (-1, 3) \cup \emptyset\] \[= (-1, 3)\]The intersection of all conditions (since condition 1 is always true) is $k \in (-1, 3)$. This means $-1 < k < 3$.
Conclusion
For both roots of the equation $x^2 - 2kx + k^2 - 4 = 0$ to lie strictly between $-3$ and $5$, the parameter $k$ must satisfy the inequality $-1 < k < 3$.
Comparing this result with the given options:
- $-2 < k < 2$
- $-5 < k < 3$
- $-3 < k < 5$
- $-1 < k < 3$
The range $-1 < k < 3$ matches the fourth option.
| Condition | Inequality | Range for $k$ |
|---|---|---|
| Discriminant $D \ge 0$ | $16 \ge 0$ | $\mathbb{R}$ (All real numbers) |
| $f(-3) > 0$ | $k^2 + 6k + 5 > 0$ | $(-\infty, -5) \cup (-1, \infty)$ |
| $f(5) > 0$ | $k^2 - 10k + 21 > 0$ | $(-\infty, 3) \cup (7, \infty)$ |
| Axis of symmetry $-3 < k < 5$ | $-3 < k < 5$ | $(-3, 5)$ |
The final range is the intersection of all these ranges: $(-1, 3)$.
Revision Table: Key Concepts for Quadratic Root Location
| Concept | Explanation | Relevance to Problem |
|---|---|---|
| Quadratic Equation $ax^2+bx+c=0$ | An equation of degree 2. Roots are values of $x$ where the equation is satisfied. | The given equation is a quadratic equation in $x$ with parameter $k$. |
| Roots of a Quadratic Equation | Given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$. Real roots exist if $D \ge 0$. | We need real roots that lie in a specific interval. |
| Discriminant $D$ | $D = b^2 - 4ac$. Determines the nature of the roots (real, complex, distinct, repeated). | Must be non-negative for real roots. Here, $D=16 > 0$ always. |
| Location of Roots | Conditions relating the roots to specific values or intervals on the number line. Depends on $a, f(p), f(q)$ and the axis of symmetry. | Crucial for solving this problem, specifically the conditions for roots lying between $-3$ and $5$. |
| Axis of Symmetry | The vertical line $x = \frac{-b}{2a}$ that passes through the vertex of the parabola $y=ax^2+bx+c$. The roots are symmetric about this line. | Its position relative to $-3$ and $5$ is a key condition. |
| Inequalities | Mathematical statements comparing two values or expressions using $<, >, \le, \ge$. Solving them helps determine valid ranges for variables. | Used extensively to find the possible values of $k$ from each condition. |
Additional Information: Solving Quadratic Root Location Problems
Problems involving the location of roots of a quadratic equation are common in algebra. The approach generally involves analyzing the sign of $a$, the sign of the function $f(x)$ at specific points (like the endpoints of the interval), the discriminant, and the position of the axis of symmetry.
- Case 1: Both roots greater than a number $p$. Conditions: $D \ge 0$, $a \cdot f(p) > 0$, $\frac{-b}{2a} > p$.
- Case 2: Both roots less than a number $p$. Conditions: $D \ge 0$, $a \cdot f(p) > 0$, $\frac{-b}{2a} < p$.
- Case 3: One root greater than $p$ and the other less than $p$. Condition: $a \cdot f(p) < 0$. (Note: $D>0$ is implied by this condition if $a \ne 0$).
- Case 4: Both roots between $p$ and $q$ ($p < q$). Conditions: $D \ge 0$, $a \cdot f(p) > 0$, $a \cdot f(q) > 0$, $p < \frac{-b}{2a} < q$. This is the case in our problem.
- Case 5: Exactly one root between $p$ and $q$ ($p < q$). Condition: $f(p) \cdot f(q) < 0$.
It is important to carefully write down and solve the inequalities resulting from each condition and then find the intersection of all the valid ranges for the parameter.
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