Question

find a quadratic polynomial whose zeroes are -3+/3, -3-/3​

Answer 1

Answer:

$let \: \alpha = - 3 + \frac{1}{3} = \frac{ - 8}{3} \\ \beta = - 3 - \frac{1}{3} = \frac{ - 10}{3} \\ using \: quadrataic \: formula \: : \\ {x}^{2} - ( \alpha + \beta )x + \alpha \beta \\ {x}^{2} - ( \frac{ - 8}{3} + ( \frac{ - 10}{3} ))x + \frac{ - 8}{3} \times ( \frac{ - 10}{3} ) \\ {x}^{2} - ( \frac{ - 8 - 10}{3})x + \frac{80}{9} \\ {x}^{2} + \frac{18}{3} x + \frac{80}{9} \\ you \: can \: put \: it \: equal \: to \: 0 \: to \: get \: it \: \\ in \: non \: fractional \: form : \\ {x}^{2} + \frac{18}{3} x + \frac{80}{9} = 0 \\ \frac{9 {x}^{2} + 54x + 80}{9} = 0 \\ 9 {x}^{2} + 54x + 80 = 0$

Step-by-step explanation:

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