Question

If one zero of the quadratic polynomial \[2{x^2} - (3k + 1)x - 9\] is negative of the other. Then Find the value of ‘k’.
A.$\dfrac{{ - 1}}{6}$
B.$3$
C.$\dfrac{1}{4}$
D.$\dfrac{{ - 1}}{3}$

Answer 1

Hint- In this question we will just follow a simple approach we will just suppose the zeroes to be $\alpha ,\beta $ and then we will just calculate the $\alpha + \beta $ using the formula for the sum of zeroes of a quadratic polynomial which is $\dfrac{{ - b}}{a}$ for the general quadratic polynomial $a{x^2} + bx + c$ . After that we will calculate the $\alpha + \beta $ using the given condition $\alpha = - \beta $ i.e. one zero is negative of the other. Then we will just compare the $\alpha + \beta $ for both the cases to get the value of k.

Complete step-by-step answer:
The quadratic polynomial so given is \[2{x^2} - (3k + 1)x - 9\]
Let \[p(x) = 2{x^2} - (3k + 1)x - 9\]
Let $\alpha ,\beta $ be the zeroes of this polynomial.
We know that for a general quadratic polynomial, $a{x^2} + bx + c$ sum of zeroes, $\alpha + \beta = \dfrac{{ - b}}{a}$
For the polynomial \[2{x^2} - (3k + 1)x - 9\] sum of zeroes, $\alpha + \beta = \dfrac{{3k + 1}}{2}$
Now it is given that $\alpha = - \beta $
$ \Rightarrow \alpha + \beta = 0$
On comparing the $\alpha + \beta $ for both the cases we get,
$ \Rightarrow \dfrac{{3k + 1}}{2} = 0$
$ \Rightarrow 3k + 1 = 0$
$ \Rightarrow k = \dfrac{{ - 1}}{3}$
Hence, the value of k $ = \dfrac{{ - 1}}{3}$
$\therefore $ Option D. $\dfrac{{ - 1}}{3}$ is our correct answer.

Note- For such types of questions, just keep in mind that for any quadratic polynomial $a{x^2} + bx + c$ sum of zeroes, $\alpha + \beta = \dfrac{{ - b}}{a}$ . Also, make the conditions properly so that in comparison we will be able to get the value we want to calculate like here we took the sum of zeroes because we can compare the sum of zeroes on the basis of the condition that one zero is the negative of the other.