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  • Solve! - If one real root of the quadratic equation is a cube of the other root, then a value of k is : - Complex numbers and quadratic equations - JEE Main

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

  • Option 1)

     

    144

  • Option 2)

     

    -300

  • Option 3)

     

    100

  • Option 4)

     

    -81

Answers (1)

best_answer

 

Sum of Roots in Quadratic Equation -

\alpha +\beta = \frac{-b}{a}

- wherein

\alpha \: and\beta are root of quadratic equation

ax^{2}+bx+c=0

a,b,c\in C

 

 

Product of Roots in Quadratic Equation -

\alpha \beta = \frac{c}{a}

- wherein

\alpha \: and\ \beta are roots of quadratic equation:

ax^{2}+bx+c=0

a,b,c\in C

 

 

Let P be 1 root of quadratic equation 

\Rightarrow other root is P^{3}

Quadratic equation can be written as - 

\left ( x-P \right )\left ( x-P^{3} \right )=0

\Rightarrow x^{2}-\left ( P+P^{3} \right )x+P^{4}=0\cdots (1)

comparing equation (1) to quadratic equation 

81x^{2}+kx+256=0

x^{2}+\frac{k}{81}+\frac{256}{81}=0

P^{4}=\frac{256}{81}\Rightarrow P=\pm \frac{4}{3}

\Rightarrow \frac{k}{81}=\pm \left ( \frac{4}{3}+\frac{64}{27} \right )=\pm \frac{300}{81}

k=\pm 300

 


Option 1)

 

144

Option 2)

 

-300

Option 3)

 

100

Option 4)

 

-81

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