Question

find the zeroes of the quadratic polynomial2x^2-3√3x-6 and verify the relationship between the zeroes and the coefficient.​

Answer 1

Answer:

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Answer 2

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Zeroes \ are \ 2\sqrt3 \ and \frac{\sqrt-3}{2}}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{2x^{2}-3\sqrt3x-6}$

$\bold{2x^{2}-4\sqrt3x+\sqrt3x-6}$

$\bold{2x(x-2\sqrt3)+\sqrt3(x-2\sqrt3)}$

$\bold{(x-2\sqrt3)(2x+\sqrt3)}$

$\bold{x=2\sqrt3 \ or \frac{-\sqrt3}{2}}$

$\bold\purple{Verification}$

$\bold{2x^{2}-3\sqrt3x-6}$

Here, a=2, b=$\bold{-3\sqrt3}$ and c=6

We know, sum of zeroes=$\bold{\frac{-b}{a}}$

$\bold{2\sqrt3+(\frac{\sqrt-3}{2})}$

=$\bold{\frac{\sqrt3}{2}...1}$

$\bold{\frac{-b}{a}}$=$\bold{\frac{\sqrt3}{2}...2}$

from 1 and 2

Sum of zeroes=$\bold{\frac{-b}{a}}$

______________________________________

Also,

$\bold{2\sqrt3×(\frac{\sqrt-3}{2})}$

=$\bold{6...1}$

$\bold{\frac{c}{a}}$=$\bold{6...2}$

from 1 and 2

Product of zeroes=$\bold{\frac{c}{a}}$