How did this equation end up with the value x006 what was th

How did this equation end up with the value x=0.06? what was the process in solving for x? thanks!
4x = 0.25 (0.5-2 (0.5-x) x = 0.06

Solution

first of all we have to perform cross multiply ,then we get

4x2=0.25 (0.5-2x)2 (0.5-x)

performing square of 0.5-x ,we get

4x2=0.25(0.25+4x2-2x) (0.5-x)

multiplying the equation variables on right hand side to simplify it ,we get

4x2=0.25(0.125+2x2-x-0.25x-4x3+2x2)

or 4/0.25 x2=-4x3+4x2-1.25x+0.125

taking terms from rhs to lhs to keep x3 constant positive

4x3+12x2+1.25x-0.125=0

now this equation is not factorised properly,so we have to take trial & error method to find the solution

If we use any negative value to x we find that result will be a negative no.,hence solution will be a positive number inetger or fraction,now if we put x=1 in the equation we get 17.125 which is a positive no.,this indicates that the solution will be in between the open interval (0,,1),now but putting x=0.5,0.2,0.1,0.05 at last we find that 0.06 is the solution of the equation.

How did this equation end up with the value x=0.06? what was the process in solving for x? thanks! 4x = 0.25 (0.5-2 (0.5-x) x = 0.06 Solutionfirst of all we hav

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