Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:-
i) $2 \sqrt{2} x^{2} - 9 x + 5 \sqrt{3}$
ii) $3 \sqrt{3} x^{2} - 19 x + 10 \sqrt{3}$ .

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Solution

i) Given quadratic equation is
$2 \sqrt{2} x^{2} - 9 x + 5 \sqrt{3} = 0$
Using quadratic formula:
$α = \frac{- b + \sqrt{D}}{2 a}$ and $β = \frac{- b - \sqrt{D}}{2 a}$
where $D = b^{2} - 4 a c = (- 9)^{2} - 4 (2 \sqrt{2}) (5 \sqrt{3}) = 81 - 40 \sqrt{6} = (- v e)$
Hence, root of above quadratic equation will be imaginary.
$∴$ , no real root exists.

ii) Given the quadratic equation is
$3 \sqrt{3} x^{2} - 19 x + 10 \sqrt{3} = 0$
Discriminant $(D) = b^{2} - 4 a c$
$x = \frac{- b \pm \sqrt{D}}{2 a} = \frac{19 \pm 1}{6 \sqrt{3}}$
$α = \frac{18}{6 \sqrt{3}} =$ $\sqrt{3}$ & $β = \frac{20}{6 \sqrt{3}} = \frac{10}{3 \sqrt{3}}$
Sum of roots $= α + β = \sqrt{3} + \frac{10}{3 \sqrt{3}} = \frac{19}{3 \sqrt{3}}$
$α + β = - \frac{(- 19)}{3 \sqrt{3}} = \frac{- (Coefficient of x)}{(Coefficient of x^{2})}$

Product of roots $= α . β = \sqrt{3} . \frac{10}{3 \sqrt{3}} = \frac{10 \sqrt{3}}{3 \sqrt{3}}$
$α . β = \frac{Constant term}{Coefficient of x^{2}}$
Hence, the relationship between the zeroes and the coefficients is verified.

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