Math, asked by Mudepakasamuel123, 10 months ago

Alpha, beeta are zeroes of quadratic polynomial p(x) =x²-6x+k if alpha-beeta =2 then find value of k


amitnrw: Use (alpha-beeta)^2 + 4alphabeeta = (alpha+beeta)^2

Answers

Answered by Anonymous
8

Answer:

8

Step-by-step explanation:

Given :

α, β are zeroes of the quadratic polynomial x² - 6x + k.

Comparing x² - 6x + k with ax² + bx + k we get

  • a = 1
  • b = - 6
  • c = k

Sum of zeroes = α + β = - b/a = - ( - 6 ) / 1 = 6

Product of zeroes = αβ = c/a = k / 1 = k

Given :

α - β = 2

Using ( α - β )² + 4αβ = ( α + β )² we get

⇒ ( α - β )² + 4αβ = ( α + β )²

⇒ ( 2 )² + 4( k ) = ( 6 )²

⇒ 4 + 4k = 36

⇒ 4k = 36 - 4

⇒ 4k = 32

⇒ k = 32 / 4

⇒ k = 8

Therefore the value of k is 8.

Similar questions