Question

If α and β are zeroes of x² –(k + 6)x + 2(2k –1). find the value of k if α + β =1/2 αβ.

Answer 1

Let α  and  β are  zeroes  of  the  polynomial  = x²  - (k  +  6)x  +  2(2k  –1).

On comparing with ax²+bx+c=0
a= 1, b= -( k+6) , c = 2(2k  –1)

Sum of zeroes (α+β)= -b/a = -(-(k+6))/1
α+β= (k+6)…………....(1)

Product of zeros(α.β)= c/a = 2(2k  –1)/1
α.β= c/a = 4k -2…………(2)

Given:  (α+β) = ½(αβ )
(k+6) = ½( 4k -2)
[From eq 1 & 2]
2 (k +6 )= 4k -2
2k +12 = 4k -2
2k -4k = -2 -12
-2k = -14
k = 14/2

k = 7
Hence, the value of k is 7
HOPE THIS WILL HELP YOU...

Answer 2

Hey there !!!!



p(x) = x2 - (k + 6)x + 2(2k - 1)

Here a = 1, b = -(k + 6) and c = 2(2k - 1)

α + β = -b/a = -[-(k + 6)]/1 = k + 6

αβ = c/a = 2(2k - 1)/1 = 2(2k - 1)

Given,

α + β = 1/2* αβ

Putting the respective values of α + β and αβ, we have

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

k + 6 = 2k - 1

k = 7

So, value of k = 7


Hope this is helpful