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

Thanks! Share it with your friends!

Learn how to solve this given radical if-then problem by using basic algebra and manipulations. Step-by-step explanation by PreMath.com

Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!

Need help with solving this Math Olympiad Question? You’re in the right place!

I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at

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

#CollegeEntranceExam
#PreMath #PreMath.com #MathOlympics #HowToThinkOutsideTheBox #ThinkOutsideTheBox #HowToThinkOutsideTheBox?

How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve international math olympiad questions
international math olympiad questions and solutions
blackpenredpen
math olympics
olympics math
olympics mathematics
olympics math activities
olympics math competition
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
exponential equation
system of equations
solve system of equations
solve the equation
solve the cubic equation
cubic equation
pre math
Poh Shen Loh
find the value of x+(4/x)
Rational Equations
Rational Equation
Competitive exams

Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

Иван Пожидаев says:

PreMath, x more than 0

Education KH says:

great solution!

Job Siu says:

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

J. R. says:

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.

Christian Thomas says:

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

Christian Thomas says:

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.

MrPaulc222 says:

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.

LIFESTYLE WITH RUKHSANA says:

Nice Method

Kevin Silva says:

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

Lee Swee Huat says:

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)

242 MATH says:

very well explained, thanks so much for sharing