If one real root of the quadratic equation 81x² + kx + 256 = 0 is cube of the other root, then a val

Q: Wait, why do we use the product of the roots first? Can't we just use the sum?

That is a fantastic question! If you start with the sum of the roots, $\alpha + \alpha^3 = -\frac{k}{81}$, you have two unknowns: $\alpha$ and $k$. You cannot solve for two variables with one equation. By starting with the product, $\alpha \cdot \alpha^3 = \frac{256}{81}$, you have an equation with only one variable, $\alpha$. This is the strategic path to the solution!

Q: Why do we get $\alpha = \pm \frac{4}{3}$? Does the sign matter for $k$?

Great observation. Since $\alpha^4 = \frac{256}{81}$, taking the fourth root gives $\alpha = \pm \frac{4}{3}$. If you test $\alpha = \frac{4}{3}$, you get $k = -300$. If you test $\alpha = -\frac{4}{3}$, you get $k = 300$. Looking at the options provided, $-300$ is the only one that matches, confirming our choice of the positive root for this specific calculation.

Q: Is there a way to solve this using the discriminant?

Technically, yes, but it is a much longer path. You would set the discriminant $D = k^2 - 4(81)(256) = 0$ if the roots were equal, but here they are distinct. You could express the roots in terms of $k$ using the quadratic formula and then set one equal to the cube of the other, but that leads to a very complex algebraic expression. Vieta's formulas are the elegant, intended shortcut for JEE problems like this!

Question

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

Options:

(A) -81

(B) 100

(C) -300

(D) 144

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to dissect a beautiful problem that tests not just your algebraic skills, but your ability to see the hidden structure within a quadratic equation.

We are given the equation $81x^2 + kx + 256 = 0$, and we are told that one root is the cube of the other. At first glance, this might seem like a daunting task—how do we find $k$ without knowing the roots?

The secret lies in the profound symmetry of quadratic equations, specifically through the lens of **Vieta's formulas**.

### The Power of Vieta's Formulas

Imagine you are standing before this equation. You have the coefficients $a = 81$, $b = k$, and $c = 256$.

The problem gives us a golden key: the roots are related. Let us define our first root as $\alpha$. Consequently, the second root is $\alpha^3$.

Now, instead of panicking about finding the roots directly, we turn to the most powerful tool in our arsenal: the relationship between roots and coefficients. We know that for any quadratic equation $ax^2 + bx + c = 0$, the product of the roots is given by:

$$ \alpha \cdot \alpha^3 = \frac{c}{a} $$

Applying this to our equation, we get:

$$ \alpha^4 = \frac{256}{81} $$

### The Hunt for $\alpha$

Now, look at the numbers. We have $\alpha^4 = \frac{256}{81}$. If you have been practicing your powers, you will immediately recognize that $256 = 4^4$ and $81 = 3^4$.

This is no coincidence! It is the problem setter rewarding your preparation. Thus, we have:

$$ \alpha^4 = \left( \frac{4}{3} \right)^4 $$

Taking the fourth root on both sides, we find $\alpha = \pm \frac{4}{3}$. We have successfully cracked the code of the roots!

### The Final Bridge to $k$

With the value of $\alpha$ in our hands, we can now hunt for $k$. Where does $k$ hide? It is the coefficient of the linear term, which means it is tied to the sum of the roots.

Vieta's formula tells us that the sum of the roots is $-\frac{b}{a}$. So:

$$ \alpha + \alpha^3 = -\frac{k}{81} $$

Let us substitute $\alpha = \frac{4}{3}$ into this expression. To make the calculation smoother, we factor out $\alpha$:

$$ \alpha(1 + \alpha^2) = -\frac{k}{81} $$

Substituting $\alpha = \frac{4}{3}$ and $\alpha^2 = \frac{16}{9}$, we get:

$$ \frac{4}{3} \left( 1 + \frac{16}{9} \right) = -\frac{k}{81} $$

Inside the bracket, $1 + \frac{16}{9}$ becomes $\frac{25}{9}$. Now we have:

$$ \frac{4}{3} \cdot \frac{25}{9} = -\frac{k}{81} \Rightarrow \frac{100}{27} = -\frac{k}{81} $$

Finally, multiplying both sides by $-81$:

$$ k = -\frac{100}{27} \cdot 81 $$

Since $81 = 27 \cdot 3$, this simplifies to $k = -100 \cdot 3$.

### Final Result

We have arrived at our destination! The value of $k$ is **$-300$**.

(Note: If we take $\alpha = -\frac{4}{3}$, we obtain $k = 300$. Depending on the context of the problem, both values are mathematically valid roots for the relationship provided.)

Animated Solution for Mathematics - Quadratic Equations: 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

Select Answer:

Visualized Solution

Identify the Quadratic Equation

Define the Roots

Apply Product of Roots Property

Simplify the Product Expression

Find the Value of $α$

Apply Sum of Roots Property

Factorize the Sum Expression

Substitute Known Values

Simplify the Bracketed Term

Multiply the Fractions

Solve for $k$

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Vieta's Formulas

The Hunt for $α$

The Final Bridge to $k$

Final Result

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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 one real root of the quadratic equation 81x2+kx+256=0 is cube of the other root, then a value of k is