Question

find the zeroes of polynomial and verify the relationship between the zeroes and cofficient 5x^2-2√5x-3​

Answer 1

HERE IS YOUR ANSWER

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

$( \sqrt{5 \: x} - 3) = 0 \\ \\ x = \frac{3}{ \sqrt{5} }$

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

Verifying relationship b/w zeroes and coefficients:

In P(x),

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

$\alpha = \frac{3}{ \sqrt{5} } \\ \\ \beta = \frac{ - 1}{ \sqrt{5} }$

$\alpha + \beta = \frac{ - b}{a} \\ \\ \frac{3}{ \sqrt{5} } + \frac{ - 1}{ \sqrt{5} } = \frac{ - ( - 2 \sqrt{5}) }{5} \\ \\ \frac{2}{ \sqrt{5} } = \frac{2 \sqrt{5} }{5} \\ \\ \frac{2 \sqrt{5} }{5} = \frac{2 \sqrt{5} }{5}$

$\alpha \beta = \frac{c}{a} \\ \\ \frac{3}{ \sqrt{5} } \times \frac{ - 1}{ \sqrt{5} } = \frac{ - 3}{5} \\ \\ \frac{ - 3}{5 } = \frac{ - 3}{ 5 } \:$

Hence, verified.

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