Question

Find the zeros of 2√3x2 – 5x + √3 and verify the relationship between the zeros
and the coefficients.​

Answer 1

Step-by-step explanation:

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

$2 { \sqrt{3}x }^{2} -(3 + 2)x + \sqrt{3 } = 0$

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

${ \sqrt{3}x }(2x - \sqrt{3} ) - 1(2x - \sqrt{3} ) = 0$

$(2x - \sqrt{3} ) \: ( \sqrt{3} x - 1)$

IF

$2x - \sqrt{3} = 0$

$x = \frac{ \sqrt{3} }{ \: 2}$

OR

$\sqrt{3} x - 1 = 0$

$x = \frac{ \: 1}{ \sqrt{3} }$

Let,

$\alpha = \frac{ \sqrt{3} }{2}$

And

$\beta = \frac{1}{ \sqrt{3} }$

So find relationship between we get.

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

$\frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{3 } } = \frac{5}{2 \sqrt{3} }$

$\frac{3 + 2} {2 \sqrt{3} } = \frac{5}{2 \sqrt{3} }$

$\frac{5}{2 \sqrt{3} } = \frac{5}{2 \sqrt{3} }$

And

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

$\frac{ \sqrt{3} }{2} \times \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{ 2\sqrt{3} }$

$\frac{1}{2} = \frac{1}{2}$

I hope it's help

Please Mark brilliant.