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PreMath, x more than 0

great solution!

Solving quatic equation of x, x=4, so answer is 5.

I squared both sides, rearranged the terms, multiplied everything by x^2 to clear the fractions, and arrived at this equation: x^4 – x^3 – 12x^2 – 4x + 16 = 0. Fortunately, this quartic isn't too difficult to solve; the roots are 1, 4, and -2 (repeated). Interestingly, both 1 and 4 give the correct answer of 5 when plugged into the x + 4/x formula, even though 1 doesn't satisfy the initial equation unless you take the negative square root. I guess this would be considered an extraneous result on that basis. -2 is clearly extraneous because it generates imaginary numbers in the original equation.

PreMath, I just red your solution after having written mine; i have to admit that your solution is very clever.

It took me one day long to find solution so I think it's worth putting it here, why not?

to make a long story short, instead of working on the expression to find, which led me nowhere, finally I just took the initial equality and squared it.

So I got (x-4/x)^2 = x + 4 + 4/x while I could see that the expression we are looking for is inserted in the RHS.

developing , i got: x^2 – 8 + 16/x^2 = x + 4 +4/x = 4 + (x+4/x) (*)

A as the LHS has been rewritten as: x^2 – 8 + 16/x^2 I had to make appear the expression to solve by writing:

x^2 + 16/x^2 = (x+4/x)^2 + 8 so that: equation (*) is written as: (x-4/x)^2 = (x+4/x)^2 – 16

Now we have: (x+4/x)^2 -16 = 4 + (x+4/x)

or: (x+4/x)^2 = 20 + (x+4/x). Now obviously we have to make a substitution like y=(x+4/x) to finally have:

y^2 – y – 20 = 0 which gives us two solutions for y: (1+/- ROOT(1+4*20))/2 which are: 10/2 = 5 and (-8/2)/2 which is negative; but as x should be positive or null the unique solution is: y=(x+4/x) = 5

Another case where it was easier to just input numbers to get the solution for x. We know X must be >0. Because a negative number inside a radical is an imaginary number, and x is in the denominator so cannot be 0.

When you have a rational variable on one side of an equation and a square root of the same variable on the other side, the variable is almost always a perfect square. Not a rule, but so common it is worth investigating before complex manipulations.

The 1st perfect square, 1, does not work. The second, 4, does.

Now that we have x the rest is simple.

I wen t a completely different route, and got the same answer!

I squared both sides for: x^2 – 8 – (16/x^2) = x + 4 + (4/x)

I then put the fractional parts aside and did a subtraction of the rest, then I subtracted one fractional part from the other. This eventually left me with

-3x^2 + 15x – 12 = 0

My coup de grace was the quadratic formula which left me with x=1 and x=4, giving an answer of 5 with either number.

I'm astonished I got it right, albeit with a very different method.

Nice Method

Tricky but amazing. Keep it up! Cheers from the Philippines

x + 4/x = 5

x^2 -5x + 4 = 0

(x-4)(x-1) = 0

x = 4 or 1

When x = 1, x – 4 / x = -3 <> sqrt(x) + 2 / sqrt(x). Thus x = 1 is not a solution and only x=4 is the solution.

The squaring process in one of the equation has introduced a wrong solution.

(-1)^2 = (1)^2 does not mean -1 = 1.

(x+4/x)^2=(x-4/x)^2+16=(sqrt(x)+2/sqrt(x))^2+18=x+4/x+20

(x+4/x)^2-(x+4/x)-20=0

x+4/x=5 or-4(rejected)

very well explained, thanks so much for sharing