If the product of the roots of the equation x² - 3kx + 2 e^(2lnk) - 1 = 0 is 7, then the roots are r

Q: Wait, why do we reject $k = -2$? The math gave me $k = \pm 2$.

That is a classic trap! Always look at the original equation before you start solving. We have a term $\ln k$ in the expression. The natural logarithm function $\ln(x)$ is only defined for $x > 0$. Since $k = -2$ would make the argument of the logarithm negative, it is outside the domain of the original equation. We must reject it.

Q: Do I really need to check the discriminant if the problem says the roots are real?

In this specific problem, the discriminant $D = k^2 + 4$ is always positive for any real $k$, so the roots are guaranteed to be real. However, in JEE Advanced, you should always verify this. If the discriminant had depended on $k$ in a way that restricted its values (e.g., $D \ge 0$ leading to $k \ge 1$), you would have had to intersect that condition with your other findings. It is a vital habit to cultivate.

Q: Is there a trick with the $e^{2\ln k}$ term? I got confused.

It is just a test of your comfort with logarithmic and exponential identities. By using the property $a \ln b = \ln b^a$, we transform $2 \ln k$ into $\ln k^2$. Then, $e^{\ln k^2}$ simplifies directly to $k^2$. It is designed to look scary to test if you panic or if you trust your fundamental properties. Stay calm, simplify, and the path forward will reveal itself.

Question

If the product of the roots of the equation $x^2 - 3kx + 2 e^{2\ln k} - 1 = 0$ is $7$, then the roots are real for $k = \dots$

Sigma Percentile · Accepted Answer

### The Art of Demystification

Welcome, fellow traveler on the path to JEE excellence. Today, we are going to dissect a problem that, at first glance, might make your heart skip a beat.

We are presented with the equation $x^2 - 3kx + 2e^{2\ln k} - 1 = 0$. It is a quadratic, yes, but that constant term—$2e^{2\ln k} - 1$—is a classic piece of psychological warfare designed to test your composure.

Let us strip away the intimidation and reveal the elegant simplicity underneath.

### Phase 1: Simplifying the Complexity

In the heat of an exam, when you see an exponential function raised to a logarithmic power, your first instinct might be to panic. Resist that.

Instead, rely on the bedrock of your mathematical training: the properties of logarithms. We know that $a \ln b = \ln b^a$.

Applying this to our term $2 \ln k$, we get $\ln k^2$. Now, the expression becomes $e^{\ln k^2}$.

Because the exponential function and the natural logarithm are inverse functions, $e^{\ln f(x)}$ simply collapses into $f(x)$. Thus, $e^{\ln k^2}$ is just $k^2$.

Suddenly, our terrifying constant term is nothing more than $2k^2 - 1$. The equation is now a friendly, standard quadratic:

$$x^2 - 3kx + (2k^2 - 1) = 0$$

### Phase 2: The Power of Vieta's Formulas

Now that we have the standard form $ax^2 + bx + c = 0$, where $a = 1$, $b = -3k$, and $c = 2k^2 - 1$, we can invoke the power of Vieta's formulas.

The problem tells us the product of the roots is $7$. We know that for any quadratic, the product of the roots is $\frac{c}{a}$.

Substituting our values, we get:

$$\frac{2k^2 - 1}{1} = 7$$

This leads us to the beautiful, simple algebraic equation: $2k^2 - 1 = 7$.

Solving this, we find $2k^2 = 8$, which means $k^2 = 4$. This gives us two potential candidates: $k = 2$ and $k = -2$.

### Phase 3: The Guardian of Reality

The problem explicitly states that the roots must be real. To ensure this, we must check the discriminant, $D = b^2 - 4ac$.

Substituting our coefficients, we get:

$$D = (-3k)^2 - 4(1)(2k^2 - 1)$$

Expanding this, we find $D = 9k^2 - 8k^2 + 4$, which simplifies to $D = k^2 + 4$.

Since $k^2$ is always non-negative for any real $k$, $k^2 + 4$ is strictly positive. This means our roots are guaranteed to be real for any real value of $k$.

We have cleared this hurdle with ease.

### Phase 4: The Final Trap

Here is where the most brilliant students sometimes stumble. We have found $k = 2$ and $k = -2$.

But we must return to the very beginning—the original equation. The term $\ln k$ exists in the problem statement.

The domain of the natural logarithm requires its argument to be strictly positive. Therefore, $k$ must be greater than $0$.

This forces us to reject $k = -2$. The only survivor, the only value that satisfies both the algebraic requirements and the domain constraints, is **$k = 2$**.

### Conclusion

This problem is a masterclass in why we must never lose sight of the big picture. We simplified the complex, applied the fundamental theorems, checked the conditions for reality, and finally, respected the domain.

Keep this discipline, and no equation will ever be too intimidating for you. You have the tools; now go forth and conquer.

Animated Solution for Mathematics - Quadratic Equations: If the product of the roots of the equation $x^{2} - 3 k x + 2 e^{2 l n k} - 1 = 0$ is $7$ , then the roots are real for $k = \dots$

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Simplify the Logarithmic Term

Rewrite the Quadratic Equation

Apply Product of Roots Condition

Formulate the Equation for $k$

Solve for $k$

Check Condition for Real Roots

Evaluate the Discriminant

Verify the Domain of Logarithm

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Art of Demystification

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

If one real root of the quadratic equation $81 x^{2} + k x + 256 = 0$ is cube of the other root, then a value of $k$ is

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $x^{2} + b x + c = 0$ , where $c < 0 < b$ , then

If $2 + i 3$ is a root of the equation $x^{2} + p x + q = 0$ , where $p$ and $q$ are real, then $(p, q) = (\dots, \dots)$ .

For the equation $3 x^{2} + p x + 3 = 0, p > 0$ , if one of the root is square of the other, then $p$ is equal to

Let $α, β$ be the roots of the equation $x^{2} - 2 x + 6 = 0$ and $\frac{1}{α ^{2}} + 1, \frac{1}{β ^{2}} + 1$ be the roots of the equation $x^{2} + a x + b = 0$ . Then the roots of the equation $x^{2} - (a + b - 2) x + (a + b + 2) = 0$ are :

Let $λ \neq = 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation :

If $α, β$ , where $α < β$ , are the roots of the equation $λ x^{2} - (λ + 3) x + 3 = 0$ such that $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$ , then the sum of all possible values of $λ$ is

If $a$ and $b$ are the roots of the equation $x^{2} - 7 x - 1 = 0$ , then the value of $\frac{a ^{21} + b ^{21} + a ^{17} + b ^{17}}{a ^{19} + b ^{19}}$ is equal to

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Let $p$ and $q$ be real numbers such that $p \neq = 0, p^{3} \neq = - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $α / β$ and $β / α$ as its roots is

Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the product of the roots of the equation x2−3kx+2e2lnk−1=0 is 7, then the roots are real for k=…