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Quadratic Equation and Inequalities question

2019 · 11 Jan · Shift 1 · Q26
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Quadratic Equation and Inequalities question

2019 · 11 Jan · Shift 1 · Q26

JEE MainMathematicsQuadratic Equation and InequalitiesMCQ+4 / −1
If one real root of the quadratic equation 81x2 + kx + 256 = 0 is cube of the other root, then a value of k is
  1. A
    −-− 81
  2. B
    −-− 300
  3. C
    100
  4. D
    144
View written solutionFree

Correct answer: B

  1. Let the roots be related as cube and number

Suppose the two real roots are rrr and r3r^3r3.

For the quadratic equation 81x2+kx+256=0,81x^2+kx+256=0,81x2+kx+256=0, by Vieta's formulas:

  • Sum of roots =−k81= -\dfrac{k}{81}=−81k​
  • Product of roots =25681= \dfrac{256}{81}=81256​
  1. Use the product of roots

Since the roots are rrr and r3r^3r3, their product is r⋅r3=r4.r\cdot r^3=r^4.r⋅r3=r4. Hence, r4=25681.r^4=\frac{256}{81}.r4=81256​. So, r=±(25681)1/4=±43.r=\pm \left(\frac{256}{81}\right)^{1/4}=\pm \frac{4}{3}.r=±(81256​)1/4=±34​.

  1. Find the corresponding pair of roots
  • If r=43r=\frac{4}{3}r=34​, then r3=6427r^3=\frac{64}{27}r3=2764​.
  • If r=−43r=-\frac{4}{3}r=−34​, then r3=−6427r^3=-\frac{64}{27}r3=−2764​.
  1. Use the sum of roots

Case 1: r=43r=\frac{4}{3}r=34​

Sum of roots: 43+6427=36+6427=10027.\frac{4}{3}+\frac{64}{27}=\frac{36+64}{27}=\frac{100}{27}.34​+2764​=2736+64​=27100​. Thus, −k81=10027-\frac{k}{81}=\frac{100}{27}−81k​=27100​ k=−81⋅10027=−300.k=-81\cdot \frac{100}{27}=-300.k=−81⋅27100​=−300.

Case 2: r=−43r=-\frac{4}{3}r=−34​

Sum of roots: −43−6427=−36+6427=−10027.-\frac{4}{3}-\frac{64}{27}= -\frac{36+64}{27}= -\frac{100}{27}.−34​−2764​=−2736+64​=−27100​. Thus, −k81=−10027-\frac{k}{81}=-\frac{100}{27}−81k​=−27100​ k=300.k=300.k=300.

  1. Match with the given options

Possible values of kkk are k=±300.k=\pm 300.k=±300. Among the given options, only k=−300k=-300k=−300 is present.

  1. Final answer

Therefore, the correct option is B.

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