Question

the value of k,so that the sum and product of roots of 2x^2+(k-3)x+3k-5=0 are equal is

Answer 1

Answer:

Step-by-step explanation:

Given that,

Let p (x) = 2x² + (k -3)x + (3k-5) = 0

Now comparing with ax²+ bx+c = 0 ,we have

a = 2 , b = ( k- 3 ) and c = (3k - 5).

Given condition is The sum and product of roots of p (x) are equal.

So, The sum of the roots = - b / a = - ( k - 3) / 2 and

The product of the roots = c / a = (3k - 5 ) / 2

∴ - ( k - 3 ) / 2 = (3k -5) / 2

⇒( - k + 2 ) / 2 = (3k -5) / 2 ( - × + = - ; - × - = + )

⇒2 ( 3k- 5 ) = 2 ( -k + 2 )

⇒ 6k - 10 = - 2k + 4 ( ∵ + × - = - )

⇒ 6k + 2k = 10 + 4 (∵ transposing )

⇒ 8k = 14

⇒ k = 14 / 8 (∵transposing )

⇒ k = 7 / 4 is the answer.