Question

If one zeroes of the quadratic polynomial \[p\left( x \right) = 4{x^2} - 8xk - 9\] is negative of the other, then find the value of \[k\].

Answer 1

Hint: In this question, we will proceed by considering one of the zeroes of the given polynomial as \[q\]then the zero will become as \[ - q\] since, it has given that the other zero or root is the negative value of the given zero or root. Then we will consider the sum of the zeroes of the polynomial to find the required value. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
The polynomial is \[p\left( x \right) = 4{x^2} - 8xk - 9\].
Let one of the zeros or roots of the given polynomial be \[q\].
Since the other zero or root of the polynomial is the negative value of the other, the other zero of the root is given by \[ - q\].
We know that for a quadratic equation \[a{x^2} + bx + c = 0\], the sum of the roots is given by \[\dfrac{{ - b}}{a}\] and the product of the roots is given by \[\dfrac{c}{a}\].
Consider the sum of the zeros or roots of the given polynomial \[p\left( x \right) = 4{x^2} - 8xk - 9\].
\[
   \Rightarrow q + \left( { - q} \right) = \dfrac{{ - \left( { - 8k} \right)}}{4} \\
   \Rightarrow q - q = \dfrac{{8k}}{4} \\
   \Rightarrow 0 = 2k \\
  \therefore k = 0 \\
\]
Thus, the value of \[k\] is 0.

Note: In any quadratic polynomial \[p\left( x \right)a{x^2} + bx + c\], the sum of zeroes is equal to the negative of the ratio of coefficient of \[x\] to the coefficient of \[{x^2}\] i.e., \[\dfrac{{ - b}}{a}\]. And the product of the zeroes or roots of the quadratic polynomial \[p\left( x \right)a{x^2} + bx + c\] is equal to the ratio of coefficient of the constant to the coefficient of \[{x^2}\] i.e., \[\dfrac{c}{a}\].