If the sum of the roots of the equation x²-(k+6)x+2(2k-1)=0 is equal to half of their product, then k =
(a)6
(b)7
(c)1
(d)5
Answers
SOLUTION :
Option (b) is correct : 7
Given : x² - (k + 6)x + 2 (2k - 1) = 0
On comparing the given equation with ax² + bx + c = 0
Here, a = 1 , b = - (k + 6) , c = 2 (2k - 1)
Sum of roots = - b/a
Sum of roots = - - (k + 6) /1
Sum of roots = k + 6
Product of roots = c/a
Product of roots = 2 (2k - 1)
Product of roots = 4k - 2
Given : sum of roots = ½ Product of roots
k + 6 = ½ (4k - 2 )
2(k + 6) = 4k - 2
2k + 12 = 4k - 2
2k - 4k = - 2 - 12
-2k = - 14
2k = 14
k = 14/2
k = 7
Hence, the value of k is 7 .
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Step-by-step explanation:
x² - (k+6)x + 2(2k+1) = 0
By comparing it with ax²+bx+c = 0,we get
a = 1 , b = -(k+6) , c = 2(2k+1)
Sum of zeroes = -b/a
= -[-(k+6)]/1
= k+6
Product of zeroes = c/a
= 2(2k+1)/1
= 2(2k+1)
As per problem,
Sum of zeroes = ½ × product of zeroes
k+6 = ½ × 2(2k+1)
k+6 = 2k+1
2k-k = 6-1
k = 5