Question

Find the value of $k$ such that the polynomial $x^2-(k+6)x+2(2k-1)$has a sum of its zeroes equal to half of their product.

Answer 1

Find the value of $k$

Given: $x^2–(k+6)x+2(2k-1)$

We know that if $\alpha ,\beta$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ then $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

Now, on comparing the given equation with the standard equation, we get; $a=1$, $b=-\left(k+6\right)$ and $c=2\left(2k-1\right)$

Given that the sum of the zeros is equal to half of the product.

$$\begin{gathered} \therefore \left(-\frac{b}{a}\right)=\frac{1}{2}\times \frac{c}{a} \\ \Rightarrow -2b=c \\ \Rightarrow 2\left(k+6\right)=2\left(2k-1\right) \\ \Rightarrow 2k+12=4k-2 \\ \Rightarrow 2k=14 \\ \Rightarrow k=7 \end{gathered}$$

Hence, the required value of $k=7$.