4y×y-15 find the zeroes of the …

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}