Question

The value of k such that polynomial x^2–(k+6)x + (2k+1) has sum of the zeroes as half of their product is :​

Answer 1

Answer :-

x² - (k+6)x + 2(2k-1)

As sum of zeros

= -b/a

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

= k+6

Product of zeros

= c/a

= 2(2k-1)/1

= 2(2k-1)

According to the question

Sum of zeros =1/2 Product of zeros

=> -b/a = 1/2 × c/a

=> k + 6 = 1/2 × 2(2k -1)

=> k + 6 = 2k -1

=> 6 + 1 = 2k - k

=> 7 = k

So Value of k = 7

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Answer 2

Answer:

x²-(k+6) x+(2k+1) =0

Comparing the above equation with ax²+bx+c=0

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

therefore,

alpha + beta=-b/a

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

and alpha×beta=c/a

=2k+1........ (2)

But it is given that sum of the zeroes are half of their product.

Since, from this given condition,

alpha +beta=alpha×beta/2

From (1) and (2),

k+6= 2k+1/2

2(k+6) =2k+1

2k+12=2k+1

12-1=2k-2k

11=0