Question

if the zeros of a polynomial p(x)=ax²+bx+c,are k+1/k and k+2/k+1,then find the value of a+b+c​

Answer 1

Step-by-step explanation:

Let a and Bare 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 (a+ß)= -b/a = -(-(k+6))/1

a+B= (k+6). .(1)

Product of zeros(a.B)= c/a = 2(2k -1)/1

a.B= c/a = 4k -2 .(2)

Given: (a+b) = 1/2(aß ) (k+6) = 2( 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 k =>7

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