Mathematical Olympiad | Solve and Check the Radical Rational equation | Math Olympiad Training
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- เผยแพร่เมื่อ 8 ก.พ. 2025
- Learn how to Solve and Check the Radical Rational equation. Step-by-step tutorial 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!
Step-by-step tutorial by PreMath.com
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Mathematical Olympiad | Solve and Check the Radical Rational equation | Math Olympiad Training
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Calm and perfect step by step resolution, as usual. Thanks, professor!
You are very welcome.
So nice of you.
Thank you for your feedback! Cheers!
You are awesome, CRMF 😀
Consistent with my talent for finding the more difficult approach:
I set P=(x-6)/(x+10) and 1/P = (x+10)/(x-6). It took more time but I did finally get the correct answer. That was a clever problem!
Thank you for your nice feedback! Cheers!
You are awesome. Keep it up 😀
Lol. But you got there
love this radical rational equation question, your step-by-step solution is awesome bro, 242 math here
Glad to hear that!
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You are awesome 😀
Beautiful problem. It got under my skin and I spent way more time on it than I will ever admit. It’s remedial math time for a retiree.
btw- The starting equations are a circle centered at 0,0 with radius sqrt(7) and and a cubic that looks roughly like a bell curve riding on a baseline y=-x. The two solutions are at the intersections of circle and bell.
Another way to solve is to let u=x+2 which gives sqrt((u-8)/(u+8)) etc. You end up with u=+or-17. I always look forward to the daily PreMath puzzle. Thanks!
You are very welcome.
Keep watching
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You are awesome, William 😀
Love and prayers from, Arizona, USA!
I did the same substitution and the cancellation was very satisfying!
Nice a task as always! I have solved it in the same way. Thank you, Professor! My congratulations on the Easter Holiday
May this Easter bring PEACE to Ukraine!
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You are awesome, Anatoliy. Stay safe and blessed 😀
Love and prayers from the USA!
Actually, youre my favourite channel on TH-cam
Your way of method is amazing
I like your method as always. Alternatively you could simply square both sides which will remove all radicals in one shot leaving a quadratic equation in terms of x right away. The two radiants being reciprocal s made the usually nasty middle term simply equal to two and made the rest of the numbers relatively smaller.
Cheers!
Very nice solution👍. Thank you teacher 🙏.
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Glad to hear that!
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You are awesome 😀
Love and prayers from the USA!
Thanks for video. Good luck sir!!!!!!
Nicely and super!!!!!!!!!
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You are awesome 😀
Love and prayers from the USA!
Did this question with ease thank you sir
.
Excellent!
Glad to hear that!
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You are awesome, Soumik 😀
very well done, thanks for sharing the solution to this radical rational equation
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So nice of you.
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You are awesome, my dear friend 😀
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal... '' '
this was very well explained, thanks for sharing bro
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So nice of you.
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You are awesome, 242 😀
Great Video!
Glad you enjoyed it
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You are awesome, Sana dear 😀
perfect solution
👌very nice explanation
Another solution method would be to first square the equation, so that you have (x-6)/(x+10)+(x+10)/(x-6)+2=1156/225. This can then be simplified to 256/(x^2+4x-60)=256/225 so that x^2+4x-60=225 must be true. And that quadratic equation has the solutions x=15 and x=-19.
Great
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You are awesome, Fretan 😀
Simple one. Thanks
-19 won't make sense
I did it by squaring both side of the original equation, and after simplification, arrived x^2 + 4x - 285 = 0, from there got the same final answers.
Professor, it is allowed to use calculator in such olympiad? By the way, congratulations for the resolution!!
very nice question
So nice of you, Nico.
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You are awesome 😀
Sir can you make a video on proof of why
If ax^a = bx^b
then a = b
Great question to ask! Pretty soon.
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You are awesome 😀
Yay! Got the same.
Bravo
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You are awesome, Saitama. Keep rocking 😀
Thnku
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You are awesome, Pranav 😀
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This equation also can be solved pretty quickly without a substitution. Just work through the algebra directly, eliminating the radicals, to end up with a huge but fun quadratic equation: 256x^2 + 1024x - 72960 = 0. This reduces to x^2 + 4x - 285 = 0, which is easy to factor as (x + 19)(x - 15).
Great
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You are awesome, JR 😀
Thanks for ur reply.
I definitely inform you as per your later example
We want numerators and denominators that result in integer square roots. x = 15 works nicely since 15 + 10 = 25 and 15 - 6 = 9. That will give us (3/5) + (5/3). (9 + 25)/(5 x 3) = 34/15.
Before solving, you need to set conditions for the variable, then there is no need to retest the solution of the equation
thanks
Спасибо за видео
Nice 👍👍
So nice of you.
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You are awesome, Umesh 😀
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V nice
So nice of you.
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You are awesome, Niru 😀
jos temenan...
15 by value putting method
Please solve this problem- 5^x + 2^x = 5
From Bangladesh
easy one
Super
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You are awesome 😀
Very nice and beautiful ❤️
I got it right ....😁
Excellent!
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You are awesome, Raj 😀
Ans : x = 15. (*Easiest question )
and -19 as well
x=15 or x=-19
Excellent!
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You are awesome 😀
9×25=225.
😭
I spent an hour trying and things scattered. In the end, I found that the solution is very simple
😐
Sometimes you r giving long sum rather than IQ based sum .
I hope ,you will prefer IQ based sum more rather than this type of monotonous sum
Dear Susen, please give an example to make this problem into an IQ based sum. Just want to see the glimpse!
Thank you for your feedback! Cheers!
34÷15=2.266666666666666666667.
g=1 giorno?
19,-15.....ho sbagliato e' -19,15
In the Soviet school, we were taught that it is necessary to determine the range of acceptable values. Here:
(x-6)/(x+10)⩾0 ⇒ [(x-6)⩾0 AND (x+10) ⩾0] OR [(x-6)⩽0 AND (x+10)⩽0] ⇒ x⩽-10 OR x⩾6