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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Quadratic Equation

JEE Main 2019 · 11 January · Morning Shift · Mathematics · Quadratic Equation

If one real root of the quadratic equation 81 x^ 2 +k x+256=0 is...

If one real root of the quadratic equation 81x2+kx+256=081 x^{2}+k x+256=0 is cube of the other root, then a value of k\mathrm{k} is :

Options

  1. A-81
  2. B100
  3. C144
  4. D-300

Correct answer

D. -300

Step-by-step solution

Let α\alpha and β\beta be the roots of the equation, 81x2+kx+256=081 x^{2}+k x+256=0 Given (α)13=β⇒α=β3(\alpha)^{\frac{1}{3}}=\beta \Rightarrow \alpha=\beta^{3} ∵\because \quad Product of the roots =25681=\frac{256}{81} ∴(α)(β)=25681\therefore \quad(\alpha)(\beta)=\frac{256}{81} ⇒β4=(43)4⇒β=43⇒α=6427\Rightarrow \quad \beta^{4}=\left(\frac{4}{3}\right)^{4} \Rightarrow \beta=\frac{4}{3} \Rightarrow \alpha=\frac{64}{27} ∵\because Sum of the roots =−k81=-\frac{k}{81} ∴α+β=−k81⇒43+6427=−k81\therefore \quad \alpha+\beta=-\frac{k}{81} \Rightarrow \frac{4}{3}+\frac{64}{27}=-\frac{k}{81} ⇒k=−300\Rightarrow \quad k=-300
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