SATHEE: Theory of Equations 1 Question 7

Theory of Equations 1 Question 7

8. 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

(2019 Main, 11 Jan I)

(a) 100

(b) 144

(d) -300

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Answer:

Correct Answer: 8. (d)

Solution:

Given quadratic equation is

$81 x^{2} + k x + 256 = 0$

Let one root be $α$ , then other is $α^{3}$ .

Now, $α + α^{3} = - \frac{k}{81}$ and $α \cdot α^{3} = \frac{256}{81}$

$[∵$ for $a x^{2} + b x + c = 0$ , sum of roots $= - \frac{b}{a}$

and product of roots $= \frac{c}{a}$ ]

$\begin{aligned} \Rightarrow α^{4} = {\frac{4}{3}}^{4} \Rightarrow α = \pm \frac{4}{3} \\ ∴ k = - 81 (α + α^{3}) \\ = - 81 α (1 + α^{2}) \\ = - 81 \pm \frac{4}{3} 1 + \frac{16}{9} = \pm 300 \end{aligned}$

Theory of Equations 1 Question 7

8. 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

Answer:

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Theory of Equations 1 Question 7

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

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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8. 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