0 Deepak BoraNewbie Asked: July 24, 20202020-07-24T14:55:05+05:30 2020-07-24T14:55:05+05:30In: CBSE Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 0 NCERT / CBSE QUESTION NO 1 (ii) CHAPTER 2.2 quadratic polynomials Share Facebook 1 Answer Voted Oldest Recent Deepak Bora Newbie 2020-07-24T15:14:48+05:30Added an answer on July 24, 2020 at 3:14 pm 4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1) Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2) Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2) Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 ) 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back ... Find the values of k for which the pair of linear equations kx + 3y = k-2 and ... . Solve for x and y: 0.4x-1.5y = 6.5, 0.3x-0.2y=0.9. Solve graphically the system of linear equations 4x-5y +16=0 and 2x+y-6 = 0. Determine the vertices of the ... If the remainder on division of x3 + 2x2 + kx +3 by x – 3 is 21, ...
4s2–2s–2s+1
= 2s(2s–1)–1(2s-1)
= (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )