Question

Pls solve question number 31

Answer 1

Answer:

$5 \sqrt{5} {x}^{2} + 20x + 3 \sqrt{5} \\ d = {b}^{2} - 4ac \\ d = 400 - 300 \\ d = 100 \\ \\ \alpha = \frac{ - b + \sqrt{d} }{2a} \\ \\ \alpha = \frac{ - 20 + \sqrt{100} }{2 \times 5 \sqrt{5} } \\ \\ \alpha = \frac{ - 10}{10 \sqrt{5} } \\ \\ \alpha = \frac{ - 1}{ \sqrt{5} } \\ \\ \beta = \frac{ - b - \sqrt{d} }{2a} \\ \\ \beta = \frac{ - 20 - 10}{2 \times 5 \sqrt{5} } \\ \\ \beta = \frac{ - 30}{10 \sqrt{5} } \\ \\ \beta = \frac{ - 3}{ \sqrt{5} } \\ \\ \\ \alpha = \frac{ - 1}{ \sqrt{5} } \: \: \: \: \: \: \: \: \: \: \beta = \frac{ - 3}{ \sqrt{5} }$

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