Math, asked by KTDBROS, 1 year ago

find all zeroes of

\sqrt{3} x^{2} +10x+7\sqrt{3}


KTDBROS: snd

Answers

Answered by TheCommando
18

Zeroes of p(x) are given by p(x) = 0

 p(x) = \sqrt{3} x^{2} +10x+7\sqrt{3} = 0

 \implies \sqrt{3} {x}^{2} + 3x + 7x + 7\sqrt{3} = 0

 \implies \sqrt{3}x (x + \sqrt{3}) + 7(x + \sqrt{3}) = 0

 \implies (\sqrt{3}x + 7) (x + \sqrt{3}) = 0

 \implies (\sqrt{3}x + 7) = 0 ; ( x + \sqrt{3} ) =0

 x = \dfrac {-7}{\sqrt{3}}

 x = -\sqrt{3}

So, Zeroes of p(x) are \dfrac {-7}{\sqrt{3}} , -\sqrt{3}.

Answered by Anonymous
0

HeRe Is Your Ans ⤵

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 { \sqrt{3x} }^{2}  + 10x + 7 \sqrt{3}  = 0 \\   \\  { \sqrt{3x} }^{2}  + 3x + 7x + 7 \sqrt{3}  = 0 \\  \\  \sqrt{3x} (x +  \sqrt{3} ) + 7(x +  \sqrt{3} ) \\ \\  \\  hence \: . \: x =  \frac{ - 7}{ \sqrt{3} }  \:  \: and  \:  \: \: x =  -  \sqrt{3}

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