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.
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!
May this Easter bring PEACE to Ukraine! Thank you for your feedback! Cheers! You are awesome, Anatoliy. Stay safe and blessed 😀 Love and prayers from the USA!
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!
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!
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.
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.
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).
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
Calm and perfect step by step resolution, as usual. Thanks, professor!
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So nice of you.
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You are awesome, CRMF 😀
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.
love this radical rational equation question, your step-by-step solution is awesome bro, 242 math here
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You are awesome 😀
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
Actually, youre my favourite channel on UA-cam
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!
Thank you for your feedback! Cheers!
You are awesome, Anatoliy. Stay safe and blessed 😀
Love and prayers from the USA!
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!
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!
Your way of method is amazing
Thanks for video. Good luck sir!!!!!!
Nicely and super!!!!!!!!!
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You are awesome 😀
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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 😀
Did this question with ease thank you sir
.
Excellent!
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You are awesome, Soumik 😀
Very nice solution👍. Thank you teacher 🙏.
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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 😀
Saraha ma3andich zhar la f tssahib la f zwaj walit tangoul tawahad maynawad tawahda Matahmal... '' '
Great Video!
Glad you enjoyed it
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You are awesome, Sana dear 😀
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
👌very nice explanation
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!!
perfect solution
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.
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
very nice question
So nice of you, Nico.
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You are awesome 😀
Yay! Got the same.
Bravo
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You are awesome, Saitama. Keep rocking 😀
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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Nice 👍👍
So nice of you.
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You are awesome, Umesh 😀
Love and prayers from the USA!
Thnku
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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
V nice
So nice of you.
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Very nice and beautiful ❤️
I got it right ....😁
Excellent!
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Please solve this problem- 5^x + 2^x = 5
From Bangladesh
easy one
Super
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You are awesome 😀
15 by value putting method
jos temenan...
Ans : x = 15. (*Easiest question )
and -19 as well
x=15 or x=-19
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😭
I spent an hour trying and things scattered. In the end, I found that the solution is very simple
😐
9×25=225.
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!
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34÷15=2.266666666666666666667.
g=1 giorno?
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
19,-15.....ho sbagliato e' -19,15