Question

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

Answer 1

Answer:

Step-by-step explanation:

x² –(k + 6)x + 2(2k –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