Question

find the zeros of the polynomial F(x)=4√3x^2+5x-2√3, and verify the relationship between the zeros and its coefficients ..

Answer 1

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

For zeros,

$4x - \sqrt{3} = 0 \\ = > \: 4x = \sqrt{3} \\ \: x = \frac{ \sqrt{3} }{4}$
And
$\sqrt{3}x + 2 = 0 \\ = > \sqrt{3} x = - 2 \\ = > x = \frac{2}{ \sqrt{3} }$
Now,
$4 \sqrt{3} {x}^{2} + 5 {x}^{2} - 2 \sqrt{3}$
$a = 4 \sqrt{3} \\ b = 5 \\ c = 2 \sqrt{3}$
Let,
$\alpha = \frac{ \sqrt{3} }{4} \: \: and \: \: \beta = \frac{2}{ \sqrt{3} }$
$\alpha + \beta = - \frac{b}{a} \\ = > \: \frac{ \sqrt{3} }{4} + \frac{2}{ \sqrt{3} } = \frac{ - 5}{4 \sqrt{3} } \\ = > \frac{3 - 8}{4 \sqrt{3} } = \frac{ - 5}{4 \sqrt{3} } \\ = > \frac{ - 5}{ 4 \sqrt{3} } = \frac{ - 5}{4 \sqrt{3} }$
And
$\alpha \beta = \frac{c}{a} \\ = > \frac{ \sqrt{3} }{4} \times \frac{2}{ \sqrt{3} } \\ = > \frac{1}{2}$
Hence verified.

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