Question

Find the zeroes of 4 root 3x2 +5x -2 root 3 and verify the relationship between the zeroes and coefficients of the polynomial

Answer 1

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Answer 2

Given:

$\bold{4\sqrt{3}x^2+ 5x -2 \sqrt{3} =0}$

To prove:

Find: zeroes and coefficients.

Solution:

$\Rightarrow 4\sqrt{3}x^2+ 5x -2 \sqrt{3} =0$

Compare the above value with $ax^2+ bx+c =0:$

$\Rightarrow a=4\sqrt{3} \\\\\Rightarrow b= 5 \\\\\Rightarrow c= -2\sqrt{3}$

Solve  above equation.

$\Rightarrow 4\sqrt{3}x^2+ 5x -2 \sqrt{3} =0\\\\\Rightarrow 4\sqrt{3}x^2+ (8 - 3)x -2 \sqrt{3} =0\\\\\Rightarrow 4\sqrt{3}x^2+ 8x-3x -2 \sqrt{3} =0\\\\\Rightarrow 4x(\sqrt{3}x+ 2) -\sqrt{3}(\sqrt{3}x +2) =0\\\\\Rightarrow (4x-\sqrt{3})(\sqrt{3}x +2) =0$

$\Rightarrow (\sqrt{3}x+ 2) = 0 \ \ \ or \ \ \ \ (4x-\sqrt{3}) =0 \\\\\Rightarrow \sqrt{3}x= - 2 \ \ \ or \ \ \ \ 4x=\sqrt{3}\\\\\Rightarrow x= - \frac{2}{\sqrt{3}} \ \ \ or \ \ \ \ x=\frac{\sqrt{3}}{4}$

zeroes and coefficients:

Formula:

$\bold{\alpha + \beta = \frac{-b}{a}}$

$\Rightarrow \frac{-2}{\sqrt{3}} + \frac{\sqrt{3}}{4} = \frac{-5}{4\sqrt{3}}\\\\\Rightarrow \frac{-8+3}{4\sqrt{3}} = \frac{-5}{4\sqrt{3}}\\\\\Rightarrow \frac{-5}{4\sqrt{3}} = \frac{5}{4\sqrt{3}}$

Formula:

$\bold{\alpha \cdot \beta = \frac{c}{a}}$

$\Rightarrow \frac{-2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{4} = \frac{- 2\sqrt{3}}{4\sqrt{3}} \\\\\Rightarrow \frac{-2\sqrt{3}}{4\sqrt{3}}= \frac{- 2\sqrt{3}}{4\sqrt{3}}$

Hence prove zeroes and coefficients are the polynomial.