Question

find the zero of polynomial​

Answer 1

Solution :

By using quadratic formula

$\implies\large \boxed{\sf x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

Here

$\sf a = 1 \: \: \: \: \: \: \: \: \: \: \: \: b = 5 \sqrt{5} \: \: \: \: \: \: \: \: \: \: \: \:c = 30$

Substitute values in formula

$\implies \sf x = \frac{ - 5 \sqrt{5} \pm \sqrt{ {(5 \sqrt{5}) }^{2} - 4 \times 1 \times 30} }{ 2\times 1} \\ \\ \implies \sf x = \frac{ - 5 \sqrt{5} \pm \sqrt{ 125- 120}}{2} \\ \\ \implies \sf x = \frac{ - 5 \sqrt{5} \pm \sqrt{5}}{2}$

Case 1 :

$\implies \boxed{ \sf x = \frac{ - 5 \sqrt{5} + \sqrt{5}}{2} }$

Case 2 :

$\implies \boxed{ \sf x = \frac{ - 5 \sqrt{5} - \sqrt{5}}{2}}$

Answer 2

I HOPE ITS HELP YOU

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