Question

find the value of K such that the ploynomial x^2-(k+6)x+2(2k-1) has sum of its zeroes equal to half of their product.​

Answer 1

Let α and β are the roots of given quadratic equation x² - ( k +6)x + 2(2k +1) = 0 [ you did mistake in typing of equation , I just correct it ]

Now, sum of roots = α + β = - {-( k + 6)}/1 = (k + 6)

product of roots = αβ = 2(2k + 1)/1= 2(2k + 1)

A/C to question,

sum of roots ( zeros ) = 1/2 × products of roots zeros

⇒ (k + 6) = 1/2 × 2(2k + 1)

⇒ (k + 6) = (2k + 1)

⇒ k + 6 = 2k + 1

⇒ k = 5

Hence, k = 5

Answer 2

Answer:

K=7

Step-by-step explanation:

x² - (k+6)x + 2(2k-1)

Sum of zeros

= -b/a

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

= k+6

Now, Product of zeros

= c/a

= 2(2k-1)/1

= 2(2k-1)

ATQ,

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

=> k = 7