Question

To factorise m^2-4m+33 solve it

Answer 1

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$\red{\bold{\underline{\underline{❥Question᎓}}}}$To factorise m^2-4m+33

$\huge\tt{\boxed{\overbrace{\underbrace{\blue{Answer</p><p> }}}}}$

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${m}^{2} - 4 m + 33$

Here we use quadratic formula to find out the roots:-

Here ,a=1 ,b=-4 & c=33

$⟹m = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}$

$⟹ m= \frac{ - ( - 4)±\sqrt{ {( - 4)}^{2} - 4(1)(33)} }{2(1)}$

$⟹m= \frac{4± \sqrt{16 - 132} }{2}$

$⟹m = \frac{4± \sqrt{ - 116} }{2} = \frac{4±2 \sqrt{ - 29} }{2}$

$⟹m = \frac{2(2± \sqrt{ - 29} )}{2} = 2± \sqrt{29i}$

Here is your answer

i(we say iota is placed of -ve sign whenever inside the root value so we simply replace negative sign by placing iota instead of negative sign inside roots)

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Answer 2

$\huge \tt{Answer}$

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$\tt{m²-4m+33}$

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$\tt{By \: Quadratic \: formula,}$

$\tt{x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{2a}}$

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$\boxed{\tt{Here, \: x = m, a = 1, b = -4, c = 33}}$

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$\tt{m= \frac{ - ( - 4)±\sqrt{ {( - 4)}^{2} - 4(1)(33)} }{2(1)}}$

$\tt{m= \frac{4± \sqrt{16 - 132} }{2}}$

$\tt{m = \frac{4± \sqrt{ - 116} }{2}}$

$\tt{ m = \frac{4±2 \sqrt{ - 29} }{2}}$

$\tt{m = \frac{\cancel{2}(2± \sqrt{ - 29} )}{\cancel{2}}}$

$\tt{m = 2 ± \sqrt{ - 29}}$

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$\red{\sf Therefore, \: value \: of \: m \: is \: 2+ \sqrt{ - 29} \: or \: 2- \sqrt{ - 29}}$