Question-56384

Question Number 56384 by Gulay last updated on 15/Mar/19

Commented by Gulay last updated on 15/Mar/19

sir plz help me

Commented by mr W last updated on 16/Mar/19

α, β = r o o t s

α = \sqrt{10} + 4

α β = 6

\Rightarrow β = \frac{6}{\sqrt{10} + 4} = \frac{6 (4 - \sqrt{10})}{4^{2} - 10} = 4 - \sqrt{10}

C = - (α + β) = - (\sqrt{10} + 4 + 4 - \sqrt{10}) = - 8

Answered by MJS last updated on 15/Mar/19

x^{2} + c x + 6 = 0

x = - \frac{c}{2} \pm \frac{\sqrt{c^{2} - 24}}{2}

- \frac{c}{2} = 4 \Rightarrow c = - 8

\frac{\sqrt{c^{2} - 24}}{2} = \sqrt{10} \Rightarrow c = \pm 8

\Rightarrow c = - 8

\Rightarrow the other solution is 4 - \sqrt{10}

(x - x_{1}) (x - x_{2}) = (x - 4 - \sqrt{10}) (x - 4 + \sqrt{10}) =

= x^{2} - 8 x + 6

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