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`sqrt(1-2x)-sqrt(16-x) = 5`
I will rearrange the above equation as,
`sqrt(1-2x) = 5 -sqrt(16-x)`
Taking the squared value of both sides,
`(sqrt(1-2x))^2 = (5 - sqrt(16-x))^2`
`(1-2x)= 25 - 10sqrt(16-x)+(16-x)`
Separating x terms and sqrt terms on to either side.
`10sqrt(16-x) = 25+16-1+2x-x`
`10sqrt(16-x) = x+40`
Again taking the squared value of bothe sides,
`(10sqrt(16-x))^2 = (x+40)^2`
`100(16-x) = x^2+80x+1600`
This gives a quadratic equation as below,
`x^2+180 = 0`
`x(x+180) = 0`
Therefore `x = 0 ` or `x =-180` .
Now try to substitute these answers in the original equation and check whether these satisfy it.
`x = 0,`
`LHS = sqrt(1)-sqrt(16) = 1 -4 = -3`
Therefore , `LHS != RHS` ,
`x = 0` is not a solution
`x = -180`
`LHS = sqrt(1+360) - sqrt(16+180)`
`= sqrt(361) -sqrt(196)`
`= 19 - 14`
Therefore x =-180 is the correct answer.
The answer is x = -180.
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