Problem 16
Question
If the product of roots of the equation \(x^{2}-\) \(3 k x+2 e^{2 \log k}-1=0\) is 7, then its roots will be a real when [IIT - 1984] (a) \(k=1\) (b) \(k=2\) (c) \(k=3\) (d) none of these
Step-by-Step Solution
Verified Answer
For real roots, the value of \(k\) must be 2 (option b).
1Step 1: Identify the Quadratic Equation
The given quadratic equation is \(x^2 - 3kx + 2e^{2\log k} - 1 = 0\). We need to analyze this equation.
2Step 2: Understand the Condition for Real Roots
For a quadratic equation \(ax^2 + bx + c = 0\) to have real roots, the discriminant must be non-negative. The discriminant \(\Delta\) is given by \(b^2 - 4ac\). If \(\Delta \geq 0\), the roots are real.
3Step 3: Use Product of Roots Condition
For the equation \(ax^2 + bx + c = 0\), the product of the roots is \(\frac{c}{a}\). Here, the product is given as 7, so we have \(2e^{2\log k} - 1 = 7\). Thus, \(2e^{2\log k} = 8\), and simplifying gives \(e^{2\log k} = 4\).
4Step 4: Simplify the Exponential Expression
Since \(e^{2\log k} = k^2\), we have \(k^2 = 4\). Solving this, \(k = \pm 2\). However, as \(k\) is positive, \(k = 2\).
5Step 5: Check Discriminant for Real Roots
Substitute \(k = 2\) into the equation to ensure real roots. The equation is now \(x^2 - 6x + 7 = 0\). Calculate the discriminant: \((-6)^2 - 4(1)(7) = 36 - 28 = 8\), which is positive, confirming real roots.
Key Concepts
Understanding the DiscriminantCalculating the Product of RootsEnsuring Real Roots
Understanding the Discriminant
In a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant is a term that plays a crucial role in determining the nature of the roots. The discriminant, denoted by \(\Delta\), is calculated using the formula \(\Delta = b^2 - 4ac\). This formula is derived from the quadratic formula and helps us predict whether the roots are real or complex.
Understanding the discriminant helps in determining not just the existence but also the nature of the roots without actually solving the entire equation.
- If \(\Delta > 0\), the quadratic equation has two distinct real roots. This means the parabola intercepts the x-axis at two distinct points.
- If \(\Delta = 0\), there is exactly one real root, usually called a double root. The parabola touches the x-axis at exactly one point.
- If \(\Delta < 0\), the quadratic equation has no real roots; instead, it has two complex roots. The parabola does not touch or intersect the x-axis.
Understanding the discriminant helps in determining not just the existence but also the nature of the roots without actually solving the entire equation.
Calculating the Product of Roots
The product of the roots of a quadratic equation is given by a simple relation derived from its coefficients. For an equation \(ax^2 + bx + c = 0\), the product of the roots \((\alpha \cdot \beta)\) can be calculated as \(\frac{c}{a}\). This result comes from Viète's formulas, which relate the roots to the coefficients of the polynomial.
In our exercise, the constraint given was that the product of the roots is 7. By using the expression for the product of roots, we have \(\frac{2e^{2\log k} - 1}{1} = 7\). Solving this gives us an insight into our equation and helps simplify the problem.
Using the product of roots concept allows students to verify or derive specific conditions about the roots without directly solving for them. It is a powerful method particularly useful in exercises where certain conditions are given or assumed.
In our exercise, the constraint given was that the product of the roots is 7. By using the expression for the product of roots, we have \(\frac{2e^{2\log k} - 1}{1} = 7\). Solving this gives us an insight into our equation and helps simplify the problem.
Using the product of roots concept allows students to verify or derive specific conditions about the roots without directly solving for them. It is a powerful method particularly useful in exercises where certain conditions are given or assumed.
Ensuring Real Roots
Ensuring that a quadratic equation has real roots requires us to check more than just solving the equation. Individual exercises often demand an evaluation of provided conditions to confirm the nature of the roots. In this context, real roots occur when the discriminant \(\Delta\) is non-negative, meaning \(\Delta \geq 0\).
In our specific quadratic equation \(x^2 - 3kx + 2e^{2\log k} - 1 = 0\), to ensure real roots, we substituted the values to find the discriminant was positive. This confirmed the presence of real roots:
By systematically checking the discriminant, students can ensure they correctly verify the conditions necessary for real roots, reinforcing their understanding of the relationship between a quadratic equation's coefficients and its roots.
In our specific quadratic equation \(x^2 - 3kx + 2e^{2\log k} - 1 = 0\), to ensure real roots, we substituted the values to find the discriminant was positive. This confirmed the presence of real roots:
- We calculated \[(-6)^2 - 4(1)(7) = 36 - 28 = 8\]
- This positive value guarantees the roots are real, fulfilling the requirement \(\Delta > 0\).
By systematically checking the discriminant, students can ensure they correctly verify the conditions necessary for real roots, reinforcing their understanding of the relationship between a quadratic equation's coefficients and its roots.
Other exercises in this chapter
Problem 14
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