Question

Thanks bro and Sister




To factorise m^2-4m+33​
.
.
.
.
.
.

ʟᴏʟʟɪᴩᴏᴩ ᴅᴇ ᴅᴏ ᴋᴏɪ...☹️ ​

Answer 1

Explanation:

factorise m^2-4m+33

\huge\orange{\overbrace{\red{\underbrace\color{blue}{\underbrace\color{black}{\colorbox{lime}{{\red\:{「Answer」}}}}}}}}

「Answer」

╔════════════════════════╗

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _✍️

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

2

−4m+33

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

\bold{Here\: ,a=1\: ,b=-4 & c=33}

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

2a

−b±

b

2

−4ac

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

2(1)

−(−4)±

(−4)

2

−4(1)(33)

⟹m= \frac{4± \sqrt{16 - 132} }{2}⟹m=

2

4±

16−132

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

2

4±

−116

=

2

4±2

−29

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

2

2(2±

−29

)

=2±

29i

\bold{Here\: is\: your\: answer}Hereisyouranswer

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)

╚════════════════════════╝

нσρє ıт нєłρs yσυ

_____________________

тнαηkyσυ