Question

answer this question ​

Answer 1

Answer:

the answer is -2

Step-by-step explanation:

x²-x+2 =0

so, x²-x = x

x= 0-2 =-2

Answer 2

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.