Question

what full solution. please help​

Answer 1

Answer:

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Answer 2

Given

${x }^{2} - 4x + 5 = 0$

$\sf since \: there \: are \: no \: real \: roots ,$

$\sf x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

here

b=4

a=1

c=5

$\sf x=\dfrac{-(-4)\pm\sqrt{4^2-(4)(5)}}{2(1)}$

$\sf x=\dfrac{4\pm\sqrt{(4(-1)}}{2}$

$\sf x=\dfrac{4\pm2i}{2}$

$\sf \boxed{ \alpha = 2 + i } \\ and \\ \sf \boxed{ \beta = 2 - i }$

Note:. i = √-1

$\sf \: now \: we \: are \: asked \: to \: find \\ \dfrac{1}{ {( \alpha + 3) }^{2} } + \dfrac{1}{ {( \beta + 3)}^{2} }$

$= > \dfrac{1}{(2 + i + 3) ^{2} } + \dfrac{1}{(2 - i+ 3) ^{2} }$

$= > \dfrac{1}{ {(5 + i)}^{2} } + \dfrac{1}{(5 - i) ^{2} }$

$= > \dfrac{1}{25 + i^2 + 10 i} + \dfrac{1}{25 + i^2 - 10 i}$

substituting i^2=-1

$= > \dfrac{1}{24 - 10 i} + \dfrac{1}{24 + 10 i}$

taking LCM

$= > \dfrac{24 + 10 i+ 24 - 10 i }{ {24}^{2} - (10 i )^{2} }$

$= > \dfrac{48}{576 - 100{i}^2}$

$= > \dfrac{48}{576 - 100( - 1)}$

$= > \dfrac{48}{676}$

$= > \dfrac{12}{169}$

Therefore the option is (1)