Question

plz answer. ........​

Answer 1

Answer:

k=7

Step-by-step explanation:

here we have given the equation,

x²-(k+6)x+2(2k-1)

compare the equation with standard form of quadratic equation, ax²+bx+c=0 , so, a = 1,b=[-(k+6)]

c=2(2k-1)

we know, the relation between coefficients and roots of the quadratic polynomial,

α+β =-b/a and (α.β)=c/a

So, as we have given condition,

α+β = 1/2(α.β)

α+β = -b/a = -(-(k+6))/(1) = (k+6)

α.β = c/a = (2(2k-1))/1 =(4k-2)

α+β = 1/2(α.β)

(k+6)=1/2(4k-2)

2(k+6)=(4k-2)

2k+12 = 4k-2 , 12+2=4k-2k

14 = 2k, K = 7

Answer 2

$\huge\bigstar\mathfrak\purple{\underline{\underline{SOLUTION}}}$

$Since \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: the \: given \: polynomial. \\\ therefore \\ = > \alpha + \beta = \frac{ - [- (k + 6)]}{1} = k + 6 \\ = > and \: \alpha \beta = 2(2k - 1)$

It is given that,

$\alpha + \beta = \frac{ 1}{2} \alpha \beta \\ = > k + 6 = \frac{1}{2} [2(2k - 1)] \\ = > k + 6 = 2k - 1 \\ = > 2k - k = 6 + 1 \\ = > k = 7$

hope it helps ☺️