Math, asked by grevathip, 3 months ago

answer this question ​

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Answers

Answered by wadhwakhushi386
0

Answer:

the answer is -2

Step-by-step explanation:

x²-x+2 =0

so, x²-x = x

x= 0-2 =-2

Answered by rajat2181
1

Given equation:

 {x}^{2}  - x + 2 = 0

a = 1, b = -1, c = 2

Roots of equation:

=x =  \frac{ - b \frac{ + }{}  \sqrt{ {b}^{2}  - 4ac} }{2a}  \\

=x =  \frac{ - ( - 1) \frac{ + }{}  \sqrt{ {( - 1)}^{2}  - 4 \times 1 \times 2} }{2 \times 1}  \\

=x =  \frac{ 1 \frac{ + }{}  \sqrt{ 1  - 8} }{2 \times 1}  \\

=x =  \frac{ 1 \frac{ + }{}  \sqrt{- 7} }{2}  = \frac{ 1 \frac{ + }{}  \sqrt{7}  \: i}{2} \\

Therefore,

 \alpha  =  \frac{1 +  \sqrt{7}  \: i}{2} \\  \beta  =  \frac{1 -  \sqrt{7 \: i} }{2}

Now,

 { \alpha }^{2}  \beta  +  \alpha  { \beta }^{2}  =

( \frac{1 +  \sqrt{7}  \: i}{2} ){}^{2}( \frac{1 -  \sqrt{7}  \: i}{2})   +( \frac{1 +  \sqrt{7}  \: i}{2} ) ( \frac{1 -  \sqrt{7}  \: i}{2}){}^{2}  \\

=( \frac{1 +  \sqrt{7}  \: i}{2} )( \frac{1   {}^{2} + ( \sqrt{7} ) {}^{2} }{ {2}^{2} }) +( \frac{1 {}^{2}  +  (\sqrt{7}) {}^{2} }{2 {}^{2} } ) ( \frac{1 -  \sqrt{7}  \: i}{2})\\

=( \frac{1 +  \sqrt{7}  \: i}{2} )( \frac{1  + 7}{ {4}}) +( \frac{1  + 7 }{4 } ) ( \frac{1 -  \sqrt{7}  \: i}{2})\\

=( \frac{1 + 7}{4} ) \: ( \frac{1 +  \sqrt{7}  \: i}{2}   +  \frac{1 -  \sqrt{7}  \:i }{2} ) \\

=( \frac{8}{4} ) \: ( \frac{1 +  \sqrt{7}  \: i + 1 -  \sqrt{7} \: i }{2}  ) \\

=2\: ( \frac{2}{2}  ) = 2 \times 1 = 2 \\

Hence, option (d) 2 is correct.

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