Question

Solve this one


thanks !!

Answer 1

$\bf HELLO\:\: DEAR,$

$\bf \: (x - 1)^{4} + (x - 5)^{4} = 82 \\ \\ \bf \: [ (x - 1)^{2})]^{2}+ [(x - 5)^{2})]^{2} = 82 \\ \\ \bf \: [(x^{2} + 1 - 2x)^{2}]+ [(x^{2} + 25 - 10x)^{2}]= 82$

$\\ \\ \bf [ x^{4} + 4x^{2} + 1 - 4x^{3} - 4x + 2x^{2} + x^{4} + 100x^{2} + 625 - 20x^{3} - 500x + 50x^{2} ] = 82$

$\\ \\ \bf \: [x^{4} + 4x^{2} + 1 - 4x^{3} - 4x + 2x^{2} + x^{4} + 100x^{2} + 625 - 20x^{3} - 500x + 50x^{2} ] = 82 \\ \\ \bf 2x^{4} - 24x^{3} + 156x^{2} - 504x + 544= 0$

$\bf {x}^{4} - 12 {x}^{3} + 78 {x}^{2} - 252x + 272 = 0$

$\bf \: {x}^{4} - 12 {x}^{3} + 78 {x}^{2} - 252x + 272 = 0 \\ \\ \bf \: (x + 4)( {x}^{3} - 8 {x}^{2} + 46x - 68) = 0$

$\bf \: (x - 4)(x - 2)( {x}^{2} - 6x + 34) = 0 \\ \\ \bf \: x = 2,4,3 +_- 5i$


$\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$

Answer 2

Given Equation is (x - 1)^4 + (x - 5)^4 = 82

= > (x - 1)^2(x - 1)^2 + (x - 5)^2(x - 5)^2 = 82

= > (x^2 + 1 - 2x)(x^2 + 1 - 2x) + (x^2 + 25 - 10x)(x^2 + 25 - 10x) = 82

= > (x^4 + x^2 - 2x^3 + x^2 + 1 - 2x - 2x^3 - 2x + 4x^2) + (x^4 + 25x^2 - 10x^3 + 25x^2 + 625 - 250x - 10x^3 - 250x + 100x^2) = 82

= > (x^4 - 4x^3 + 6x^2 - 4x + 1) + (x^4 - 20x^3 + 150x^2 - 500x + 625) = 82

= > x^4 - 4x^3 + 6x^2 - 4x + 1 + x^4 - 20x^3 + 150x^2 - 500x + 625 = 82

= > 2x^4 - 24x^3 + 156x^2 - 504x + 626 = 82

= > 2x^4 - 24x^3 + 156x^2 - 504x + 544 = 0

= > 2(x^4 - 12x^3 + 78x^2 - 252x + 272) = 0

= > x^4 - 12x^3 + 78x^2 - 252x + 272 = 0

= > x^4 - 10x^3 - 2x^3 + 58x^2 + 20x^2 - 136x - 116x + 272 = 0

= > x^4 - 10x^3 + 58x^2 - 136x - 2x^3 + 20x^2 - 116x + 272 = 0

= > x(x^3 - 10x^2 + 58x - 136) - 2(x^3 - 10x^2 + 58x - 136) = 0

= > (x - 2)(x^3 - 10x^2 + 58x - 136) = 0

= > (x - 2)((x^3 - 6x^2 - 4x^2 + 34x + 24x - 136) = 0

= > (x - 2)((x(x^2 - 6x + 34) - 4(x^2 - 6x + 34)) = 0

= > (x - 2)(x - 4)(x^2 - 6x + 34) = 0

= > (x - 2)(x - 4)(x^2 - 6x + 34) = 0

(1)

= > x - 2 = 0

x = 2.


(2)

= > x - 4 = 0

x = 4


(3)

= > x^2 - 6x + 34 = 0

a =1, b = -6, c = 34

(i)

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

$= \ \textgreater \ \frac{-(6) + \sqrt{(-6)^2 - 4(1)(34)} }{2(1)}$

$= \ \textgreater \ \frac{6 + \sqrt{36 - 136}}{2}$

$= \ \textgreater \ \frac{6 + \sqrt{100}i }{2}$

$= \ \textgreater \ \frac{6 +10i}{2}$

$= \ \textgreater \ \frac{2(3 + 5i)}{2}$

$= \ \textgreater \ 3 + 5i$


(ii)

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

$= \ \textgreater \ \frac{-(-6) - \sqrt{(-6)^2 - 4(1)(34)} }{2}$

$= \ \textgreater \ \frac{6 - \sqrt{36 - 136} }{2}$

$= \ \textgreater \ \frac{6 - \sqrt{-100} }{2}$

$= \ \textgreater \ \frac{6 - 10i }{2}$

$= \ \textgreater \ \frac{2(3 - 5i)}{2}$

$= \ \textgreater \ (3 - 5i)$



Therefore the value of x = 2, 4, 3 + 5i, 3 - 5i.



Hope this helps!