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| 1. |
4y×y-15 find the zeroes of the polynomial |
| Answer» We have to find the zeroes of the quadratic polynomial 4y2 – 15 and verify the relationship between the zeroes and coefficient of polynomial.Let {tex}f(y)\\;=\\;4y^2\\;–\\;15{/tex}Compare it with the quadratic {tex}ay^2\\;+\\;by\\;+\\;c{/tex}.Here, coefficient of{tex}\\;y^2\\;=\\;4{/tex}, coefficient of y = 0 and constant term = - 15.Now {tex}4y^2\\;–\\;15\\;=\\;(2y)^2\\;–\\;(\\;\\sqrt{15})^2{/tex}= {tex}(2y\\;+\\;\\;\\sqrt{15})(2y\\;-\\;\\sqrt{15}){/tex}The zeroes of f(y) are given by {tex}f(y) = 0{/tex}⇒{tex}(2y)\\;+\\;\\;\\sqrt{15})(2y\\;-\\;\\sqrt{15}){/tex} = 0⇒ {tex}(2y)\\;+\\;\\;\\sqrt{15}){/tex} = 0 or {tex}(2y\\;-\\;\\;\\sqrt{15}){/tex} = 0⇒ {tex}2y\\;=\\;-\\;\\;\\sqrt{15}{/tex} or {tex}2y\\;=\\; \\;\\;\\sqrt{15}{/tex}⇒ {tex}\\;y\\;=\\;-\\frac{\\;\\;\\sqrt{15}}2{/tex} or {tex}\\;y\\;=\\;\\frac{\\;\\;\\sqrt{15}}2{/tex}Hence, the zeroes of the given quadratic polynomial are {tex}-\\frac{\\;\\;\\sqrt{15}}2{/tex}, {tex} \\frac{\\;\\;\\sqrt{15}}2{/tex} | |