Question

If the zeros of the polynomial $ f\left( x \right) = k{x^3} - 8{x^2} = 5 $ are alpha - beta, alpha and alpha + beta then find the value of k.

Answer 1

Hint: The given polynomial is a cubic polynomial and to find the value of k, we are going to use three relations between the zeros of the polynomial and the coefficients of the polynomial.
 $ \Rightarrow \alpha + \beta + \gamma = \dfrac{{ - b}}{a} $
 $ \Rightarrow \alpha \beta \gamma = \dfrac{{ - d}}{a} $
 $ \Rightarrow \alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a} $

Complete step-by-step answer:
In this question, we are given a cubic polynomial and we are given that its zeros are $ \alpha - \beta $ , $ \alpha $ and $ \alpha + \beta $ and we are supposed to find the value of k.
Given polynomial: $ k{x^3} - 8{x^2} = 5 $
 $ \Rightarrow k{x^3} - 8{x^2} - 5 = 0 $
Here, $ a = k,b = - 8,c = 0,d = - 5 $
Now, this is a cubic polynomial and we need to find the coefficient of $ {x^3} $ .
For finding the value of k, we are going to use the relations between the zeros of the polynomial and the coefficients of the polynomial.
Sum of zeros
First relation is that the sum of the zeros of a cubic polynomial is $ \dfrac{{ - b}}{a} $ .
 $ \Rightarrow \alpha + \beta + \gamma = \dfrac{{ - b}}{a} $
 $
   \Rightarrow \alpha + \beta + \alpha + \alpha - \beta = \dfrac{{ - b}}{a} \\
   \Rightarrow 3\alpha = \dfrac{{ - \left( { - 8} \right)}}{k} \;
  $
 $ \Rightarrow 3\alpha = \dfrac{8}{k} $
 $ \Rightarrow k = \dfrac{8}{{3\alpha }} $ - - - - - - - - - - (1)
Product of zeros
Second relation is that the product of the zeros of a cubic polynomial is $ \dfrac{{ - d}}{a} $ .
 $ \Rightarrow \alpha \beta \gamma = \dfrac{{ - d}}{a} $
 $ \Rightarrow \left( {\alpha - \beta } \right)\left( {\alpha + \beta } \right)\left( \alpha \right) = \dfrac{{ - \left( { - 5} \right)}}{k} $
Now, we know that $ \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} $ . Therefore,
 $ \Rightarrow \left( {{\alpha ^2} - {\beta ^2}} \right)\left( \alpha \right) = \dfrac{5}{k} $
Now, from equation (1), substitute the value of k.
 $ \Rightarrow \left( {{\alpha ^2} - {\beta ^2}} \right)\left( \alpha \right) = \dfrac{5}{{\dfrac{8}{{3\alpha }}}} = \dfrac{{5 \times 3\alpha }}{8} $
 $ \Rightarrow \left( {{\alpha ^2} - {\beta ^2}} \right) = \dfrac{{15}}{8} $
 $ \Rightarrow \left( {{\alpha ^2} - {\beta ^2}} \right) = \dfrac{{15}}{8} $
\[ \Rightarrow {\beta ^2} = {\alpha ^2} + \dfrac{{15}}{8}\] - - - - - - (2)
Sum of product of roots
Third relation is that the sum of product of zeros of a cubic polynomial is $ \dfrac{c}{a} $ .
 $ \Rightarrow \alpha \beta + \beta \gamma + \alpha \gamma = \dfrac{c}{a} $
 $ \Rightarrow \alpha \left( {\alpha - \beta } \right) + \left( {\alpha - \beta } \right)\left( {\alpha + \beta } \right) + \left( {\alpha + \beta } \right)\alpha = \dfrac{0}{k} $
Opening the brackets, we get
 $
   \Rightarrow \alpha \left( {\alpha - \beta } \right) + \left( {\alpha - \beta } \right)\left( {\alpha + \beta } \right) + \left( {\alpha + \beta } \right)\alpha = \dfrac{0}{k} \\
   \Rightarrow {\alpha ^2} - \alpha \beta + {\alpha ^2} - {\beta ^2} + {\alpha ^2} + \alpha \beta = 0 \\
   \Rightarrow 3{\alpha ^2} - {\beta ^2} = 0 \\
  $
Now, from equation (2), we get
 $
   \Rightarrow 3{\alpha ^2} - \left( {{\alpha ^2} + \dfrac{{15}}{8}} \right) = 0 \\
   \Rightarrow 3{\alpha ^2} - {\alpha ^2} - \dfrac{{15}}{8} = 0 \;
  $
 $ \Rightarrow 2{\alpha ^2} = \dfrac{{15}}{8} $
Now, from equation (1), $ \alpha = \dfrac{8}{{3k}} $
Therefore,
\[
   \Rightarrow 2{\left( {\dfrac{8}{{3k}}} \right)^2} = \dfrac{{15}}{8} \\
   \Rightarrow \left( {\dfrac{{64}}{{9{k^2}}}} \right) = \dfrac{{15}}{{16}} \\
   \Rightarrow {k^2} = \dfrac{{64 \times 16}}{{15 \times 9}} \\
   \Rightarrow k = \sqrt {\dfrac{{64 \times 16}}{{15 \times 9}}} \\
   \Rightarrow k = \dfrac{{8 \times 4}}{{3 \times \sqrt {15} }} \\
   \Rightarrow k = \dfrac{{32}}{{3\sqrt {15} }} \;
 \]
Hence, the value of k is \[\dfrac{{32}}{{3\sqrt {15} }}\].
So, the correct answer is “ \[\dfrac{{32}}{{3\sqrt {15} }}\]”.

Note: The relation between the zeros and the coefficients of a quadratic equation are:
Sum of zeros $ \alpha + \beta = \dfrac{{ - b}}{a} $
Product of zeros $ \alpha \beta = \dfrac{c}{a} $
Note that the equation has a number of zeros equal to the power of the highest degree term in the equation.