Question

find the roots of the equation
${x}^{2} + 5 = - 6x \: using \: formula \: metod$

Answer 1

X^2+6x+5=0

By quadratic formula x=-b+-underrootofb^2-4ac/2a

X=-6underrootof 6^2-4**5/2*1

X=-6+-4/2

X=-10/2,-2/2

X=-5,-1

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Answer 2

$\huge{\bold{\underline{\underline{\underline{Heya!!}}}}}$

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Ques:- find the roots of the equation
${x}^{2} + 5 = - 6x \: using \: formula \: metod$

$\bold{\huge{\underline{Answer:-}}}$

Given eq.

${x}^{2} + 5 = - 6x$
Transposing -6x to LHS

${x}^{2} +6x + 5 = 0$

Comparing it with std form of quad. eq

ax^2+bc+c =0

We get,

=>a=1, b =5. and. c

$as \: we \: know \: that \: \\ \\ quadratic \: formula = \\ \\ \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ \\ now \: putting \: the \: value \: of \: a \: b \: and \: c \: \: \\ in \: the \: given \: formula \\ x = \frac{ - 6 + - \sqrt{ {6}^{2} - 4.1.5} }{2.1} \\ \\ x = \frac{ - 6 + - \sqrt{36 - 20} }{2} \\ \\ x = \frac{ - 6 + - \sqrt{16} }{2} \\ \\ \\ x = \frac{ - 6 + 4}{2} \: \: \: \frac{ - 6 - 4}{2} \\ \\ x = \frac{ - 2}{2} \: \: \: \: \frac{ - 10}{2} \\ \\ \\ x = - 1 \: \: - 5$

$<b><U>So, roots of the given eqn are -1, -5</U></b>$

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