Math, asked by kakasingh1530, 5 months ago

the rots of equation 5ײ-4×5=0 are​

Answers

Answered by aryan073
1

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➡️\huge\mathfrak\pink{Solution :-}

 \:  \:  \to \underline{ \boxed{ \bf \red{the \: roots \: of \: equation \:  {5x}^{2}  - 4x + 5 = 0 \: are}}}

 \:  \:  \:  \:  \bigstar \bf{by \: using \: determinant \: form}

 \:  \to \rm{ {5x }^{2}  - 4x + 5 = 0}

 \:  \to \rm{ {b}^{2}  - 4ac}

 \:  \:  \to \rm{ {(4)}^{2}  - 4(5)(5)}

  \: \to \rm{16 - 100}

 \:  \:  \to \rm{ - 84}

 \:  \:  \bigstar\bf{by \: formula \: method}

 \:  \:  \:  \:  \to \sf{x =  \frac{ - b \pm \sqrt{ {b}^{2}  - 4ac} }{2a} }

 \:  \:  \to \rm{x =  \frac{4 \pm \sqrt{ - 84} }{2 \times 5} }

 \:  \:  \to \rm{x =  \frac{4 \pm 2\sqrt{21} }{10} }

 \:  \:  \:  \to \rm{x =  \frac{2(2 \pm \sqrt{21}) }{2 \times 5} }

 \:  \:  \to \rm{x =  \cancel   \red {\frac{2}{2} } \frac{2 \pm  \sqrt{21} }{5} }

 \:  \:  \to \rm{x =  \frac{2 \pm \sqrt{21} }{5} }

 \:  \:  \:  \:  \dashrightarrow \boxed{ \bf \pink{ \: the \: roots \: are \: x =  \frac{2 +  \sqrt{21} }{5} and \: x =  \frac{2 -  \sqrt{21} }{5} }}

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