Question

If sum of the zeroes of the polynomial x2 - kx + 8 is 6, then the value of
k will be​

Answer 1

Answer:

$therefore \: \: \: \: k = 6$

Step-by-step explanation:

${x}^{2} - kx + 8 = 0 \\ \\ sum \: of \: zeroes \: = 6 \\ \\ - \frac{coefficient \: of \: x}{ coefficient \: of \: {x}^{2} } \\ \\ = \frac{ - ( - k)}{1} = 6 \\ \\ therefore \: \: \: \: k = 6$

Answer 2

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{a \: polynomial \: {x}^{2} - kx + 8 } \\ &\sf{sum \: of \: zeroes \: = \: 6} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{ \: value \: of \: k}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Now,

We know that,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

OR

$\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

Given polynomial is

$\sf \: {x}^{2} - kx \: + 8$

$\red{\sf Sum\ of\ the\ zeroes=\dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2} } = 6}$

$:  \implies  \tt \: - \dfrac{( - k)}{1} = 6$

$:  \implies \large\boxed{ \purple{  \tt \: k \: = \: 6}}$