Question

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

Answer 1

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 .

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Solution :

Let m , n are two roots of

given Quadratic equation,

Compare x²-(k+6)x+2(2k-1)=0

with ax²+bx+c = 0, we get

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

i ) m+n = -b/a

= -[-(k+6)]

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

ii ) Product of the roots

= c/a

= 2(2k-1) ----( 2 )

According to the problem given,

m+n = ( mn/2 )

=> k+6 = [2(2k-1)]/2

=> k+6 = 2k - 1

=> 6 + 1 = 2k - k

=> 7 = k

Therefore ,

k = 7

Option (b) is correct.

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