Question

Solve :-
$(x - 1)(x - 2)(x - 3)(x - 4) = 8$

Answer 1

Hey mate !!

Here's the answer !!

( x - 1 ) ( x - 2 ) ( x - 3 ) ( x - 4 ) = 8

=> ( x ² - 2x - x + 2 ) ( x² - 4x - 3x + 12 ) = 8

=> ( x² - 3x + 2 ) ( x² - 7x + 12 ) = 8

=> x⁴ - 7x³ + 12x² - 3x³ + 21x² - 36x + 2x² - 14x + 24 = 8

Combining the like terms, we get,

=> x⁴ - 10x³ + 35x² - 50x + 24 = 8

=>  x⁴ - 10x³ + 35x² - 50x + 24 - 8 = 0

=> x⁴ - 10x³ + 35x² - 50x + 16 = 0

Solving the above equation, we get,

$x = \frac{- \sqrt{17} + 5}{2} , x = \frac{ \sqrt{17} +5}{2}$

Hope my answer helps !!

Cheers !!

Answer 2

Hey friend, Harish here.

Here is your answer. 

$(x - 1)(x - 2)(x - 3)(x - 4) = 8 \\ \\ \to \left(x^2-3x+2\right) (x^2-7x+12) = 8 \\ \\ \to x^4-10x^3+35x^2-50x+24 = 8 \\ \\ \to x^4-10x^3+35x^2-50x+16 = 0 \\ \\ \to \Bigl( \bigl( x-5 \bigr)x +2\Bigr ) \Bigl( \bigl( x-5 \bigr)x + 8 \Bigr ) = 0$ 

$\to (x^2 - 5 x + 2) (x^2 - 5 x + 8) = 0 \\ \\ \mathrm {Now, Let's \ Check \ the \ nature\ of\ roots \ of\ both\ quadratic\ equations} \\ \\ \mathrm{ In \ (x^2 - 5 x + 2) \ ; a = 1,b=-5,c=2 } \\ \\ Then,\ b^2 - 4ac = 25 - 28 = -3 \ \textless \ 0 \\ \\ \mathrm{Therefore\ it\ has\ no\ real\ roots} \\ \\ \mathrm{Then, The \ real\ value\ of\ x \ is \ the \ roots\ of\ (x^2 - 5 x + 2)}\\ \\ \mathrm{By\ Quadratic\ fromula} : \\ \\ \boxed{\bold{x=\frac{5-\sqrt{17}}{2},\:x=\frac{5+\sqrt{17}}{2}}}$
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Hope my answer is helpful to you.