Olympiad Mathematics | Learn to find the value of x+(4/x) | Math Olympiad Preparation

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Olympiad Mathematics | Learn to find the value of x+(4/x) | Math Olympiad Preparation

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[christianthomas]

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

[MrPaulc]

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.

[jiteshkumarraut]

Thanks Pre Math for this short method of solving this problem.

[kennethstevenson]

Great insight, being able to recognize common factors of irrational expressions of the difference of two squares, then squaring both sides to form the answer to the equation to find x + 4/x = 5.

[alster]

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

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing✨✨

[math]

very well explained, thanks so much for sharing

[predator]

Excellent solution 👍, thank you teacher 🙏.

[ankeshpritam]

Nice solution of this problem
Thanks 😊

[j.r.]

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.

[AdityaKumar-gvdj]

Since, x is a real number so it is possible that it can be negative so x=-2 is also a solution to this problem and so the answer x+1/x can also be equal to -4.

[murdock]

Awesome, many thanks. Try this: (x² - 4)/x = (x + 2)/√x → √x = (x - 2) → x^2 - 5x + 4 = 0 → x1 = 4, x2 = 1 🙂

[SuperYoonHo]

Thank you so much sir the answer is 5. Keep up your good work friend and i will keep suporting you;)

[thorinpalladino]

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.

[seegeeaye]

from your answer, x = 1, which is not satisfied with the given equation.
but x = 4 is.

[douglasfeather]

I did this a completely different way. Starting with 4 - 4/x = sqr(x) + 2/sqr(x) I squared both sides and then substituted y = x + 4/x. I was able to rearrange this to y^2 - y - 20 = 0. This factorises as (y+4)(y-5) = 0 which gives the solution of 5 shown in the video. It also gives a solution of -4 which I believe is missed by the video as he divides by sqr(x) + 2/sqr(x) which is zero for the solution of -4. [x + 4/x = -4 => x=-2. Substituting sqr(x) = +2i or -2i into sqr(x) + 2/sqr(x) gives zero for both values]

[satrajitghosh]

a bit of work gives
EITHER √x - 2/√x = 1
x + 4/x = (√x - 2/√x)^2 + 4 = 5
OR √x + 2/√x = 0
x + 4/x = (√x + 2/√x)^2 - 4 = -4

[miguelgnievesl]

I used another method and got two results. (X + 4/X)= 5 and (X + 4/X)= -4.

[johnwindisch]

Solved it a different way but should have checked the solutions. Results were 1, -2 and 4. Only 4 worked.

[leesweehuat]

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.