Question

2root3x2-5x+root find zero polynomial and verify the relation between the zeros and coefficients

Answer 1

Answer:

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if your question is as follows

Step-by-step explanation:

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$2 \sqrt{3} {x}^{2} - 5x + \sqrt{3} \\ 2 \sqrt{3} {x}^{2} - 2x - 3x + \sqrt{3} = 0 \\ 2x( \sqrt{3} x - 1) - \sqrt{3} ( \sqrt{3} x - 1) = 0 \\ ( \sqrt{3} x - 1)(2x - \sqrt{3} ) = 0 \\ \sqrt{3} x - 1 = 0 \: \: \: \: \: \: 2x - \sqrt{3} = 0 \\ \sqrt{3} x = 1 \: \: \: \: \: \: \: \: 2x = \sqrt{3} \\ x = \frac{1}{ \sqrt{3} } \: \: \: \: \: \: \: \: \: x = \frac{ \sqrt{3} }{2}$

sum of zeroes = -b/a = -(-5)/2√3 = 5/2√3

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

product of zeroes = c/a = √3/2√3 = 1/2

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

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