WTF? Problem from olympics? This is 7th grade problem, No log's needed.
@moonwatcher200116 сағат бұрын
❤
@HoSza120 сағат бұрын
why does he reject complex solitions? that's unreasonable. 😢
@pukuluКүн бұрын
well, if x to the 12 = 64, and x to the 8 = 16 we have a solution. x is the 4th root of 4, which is the square root of 2.
@HoSza120 сағат бұрын
suspiciously many of people don't recognize that 4¼=√2 😂
@payoo_2674Күн бұрын
8:35 |x| = 4^(1/4) x = 4^(1/4) or x = -4^(1/4) 😀
@math3starКүн бұрын
nice video
@math3starКүн бұрын
Nice math
@arrao633Күн бұрын
5*7=35. 7*8=56. 8*5=40. No algebra required. Simple thinking is all that is needed. Can be solved in less than half a minute.
@sans13313 күн бұрын
5^(x+6)=6^(x+5) ln both sides (x+6)ln5=(x+5)ln6 xln5+6ln5=xln6+5ln6 -xln6 and -6ln5 both sides xln5-xln6=(ln5-ln6)x=5ln6-6ln5 x=(5ln6-6ln5)/(ln5-ln6)
@pjaj434 күн бұрын
Worth noting that the substitution of b = 100/a is completely symmetrical , a and b are interchangeable, you can say a = 100/b and get the same pair of roots. Hence b1 = a2 and b2 = a1 without any further calculation as you know a quadratic only has two roots so they must be the same complementary pair.
@aminimam51184 күн бұрын
There are two other imaginary solutions X^3=27^2=(3^2)^3 X^3-(3^2)^3=0 (x-9)(x^2+9x+81)=0
@manivasagamrajalingam18684 күн бұрын
A=5. B=7. C=8
@dubdub66386 күн бұрын
when you have a sum and a product of two numbers, I think the simplest solition is: t² - St + P = 0 I like your channel. Bye from Italy!
@dubdub66386 күн бұрын
t²-St+P =0 where S =a+b and P = ab t²-4t+24=0 ...
@alipourzand64996 күн бұрын
Neat! What about this: sqrt(a) and sqrt(b) are the roots of (x - sqrt(a)) (x - sqrt(b)) = 0 Then using the quadratic formula to find a and b
@davidjohnsummers75636 күн бұрын
Substitute u=sqrt(a) and v=sqrt(b). Then solve for u and v. Finally go back to a and b.
@mieses-te9yl5 күн бұрын
Agree!
@MaGaO4 күн бұрын
No need: one can square "√a+√b=10" to get "a+2√(ab)+b=100" and replace √(ab) by 10.
Being a biquadratic equation it will have four solutions.
@carlosrivas20127 күн бұрын
Un ejercicio simpático. Desde el enunciado se ve la trampa..... no pueden ser reales.
@coooling077 күн бұрын
1/9
@periathiruvadi17067 күн бұрын
A-5,B-7,C-8
@AbouTaim-Lille8 күн бұрын
Supposing that the solution is in the field of real numbers. Since their multiplication has a positive sign the 2 roots have the same sign. And since their sum is positive then they are both positove numbers. And since the highest multiplcation they can assume for 2 numbers that have the sum 10 is when theyre equal to their sum devided by 2 which is 5 and we have 5x5 =25 < 150 then we deduce that the set of equations had no REAL roots.
Is it necessary to show a^m+a^n =a^m+n. You think we are stupid and you are the clever one. 4Q
@evlorenzoni10 күн бұрын
Check the video again. You are wrong.
@mohsinrazaabidi892511 күн бұрын
a 5 b 7 c 8
@ratansoren628812 күн бұрын
ABC 578
@ManojkantSamal12 күн бұрын
ab=35.......eqn1 bc=56........eqn2 ca=40.........eqn3 eqn1÷eqn2, we shall get ab/bc=35/56 a/c=5/8.....eqn4 Again eqn3÷eqn2, we shall get ca/bc=40/56 a/b=5/7......eqn5 eqn4×eqn5, we shall get (a/c)×(a/b)=(5/8)×(5/7) a^2/bc =25/56 a^2/56=25/56 (bc=56) a^2=25 a=5 Put the value of a in eqn1 5b=35 b=7 Put the value of b in eqn2 7c=56 C=8 Hence a=5, b=7,c=8
@garydetlefs609512 күн бұрын
I agree with the comments below that this problem could be easily solved by inspection. However if the numbers were larger and the factors not so obvious there is another interesting way about going about solving this problem Set up a 3x4 matrix [1,1,0,ln(35),0,1,1,ln(56),1,0,1,ln(40)] Do Gauss Jordan Elimination and obtain ln(5), ln(7), ln(8). Raise e to these values. Also realized it it would apply to negative numbers as well. Overkill for this particular problem but it is an interesting approach to these types of problems which at first glance would not seem applicable to linear algebra methods.
@fredericiovleff909712 күн бұрын
You said "without calculator", so I've made : 7x7 = 49 -> sqrt(48) ~= 6.9 sqrt(48) + sqrt(16) ~= 6.9 + 4 = 10.9 ~= 11 sqrt(48) - sqrt(16) ~= 6.9 - 4 = 2.9 ~= 3 11/3 ~= 3.7 it's pretty near from sqrt(3) + 2 ~= 3.73205080757
Пікірлер
WTF? Problem from olympics? This is 7th grade problem, No log's needed.
❤
why does he reject complex solitions? that's unreasonable. 😢
well, if x to the 12 = 64, and x to the 8 = 16 we have a solution. x is the 4th root of 4, which is the square root of 2.
suspiciously many of people don't recognize that 4¼=√2 😂
8:35 |x| = 4^(1/4) x = 4^(1/4) or x = -4^(1/4) 😀
nice video
Nice math
5*7=35. 7*8=56. 8*5=40. No algebra required. Simple thinking is all that is needed. Can be solved in less than half a minute.
5^(x+6)=6^(x+5) ln both sides (x+6)ln5=(x+5)ln6 xln5+6ln5=xln6+5ln6 -xln6 and -6ln5 both sides xln5-xln6=(ln5-ln6)x=5ln6-6ln5 x=(5ln6-6ln5)/(ln5-ln6)
Worth noting that the substitution of b = 100/a is completely symmetrical , a and b are interchangeable, you can say a = 100/b and get the same pair of roots. Hence b1 = a2 and b2 = a1 without any further calculation as you know a quadratic only has two roots so they must be the same complementary pair.
There are two other imaginary solutions X^3=27^2=(3^2)^3 X^3-(3^2)^3=0 (x-9)(x^2+9x+81)=0
A=5. B=7. C=8
when you have a sum and a product of two numbers, I think the simplest solition is: t² - St + P = 0 I like your channel. Bye from Italy!
t²-St+P =0 where S =a+b and P = ab t²-4t+24=0 ...
Neat! What about this: sqrt(a) and sqrt(b) are the roots of (x - sqrt(a)) (x - sqrt(b)) = 0 Then using the quadratic formula to find a and b
Substitute u=sqrt(a) and v=sqrt(b). Then solve for u and v. Finally go back to a and b.
Agree!
No need: one can square "√a+√b=10" to get "a+2√(ab)+b=100" and replace √(ab) by 10.
🤣🤣🤣🤣
xy+x=32 xy+y=35 So... x=y-3 Y(y-3)+y-3=32 Y²-3y+y-3-32=0 Y²-2y-35=0 D=4+140=144 Y1,y2= 2±12/2 Y1=7 , y2=-5 X1=4, x2=-8
Excelente. Gracias.
Being a biquadratic equation it will have four solutions.
Un ejercicio simpático. Desde el enunciado se ve la trampa..... no pueden ser reales.
1/9
A-5,B-7,C-8
Supposing that the solution is in the field of real numbers. Since their multiplication has a positive sign the 2 roots have the same sign. And since their sum is positive then they are both positove numbers. And since the highest multiplcation they can assume for 2 numbers that have the sum 10 is when theyre equal to their sum devided by 2 which is 5 and we have 5x5 =25 < 150 then we deduce that the set of equations had no REAL roots.
√[2x - 5] = x - 4 2x - 5 = (x - 4)^2 = x^2 - 8x + 16 x^2 - 10x + 21 = 0 (x - 3)(x - 7) = 0 3 is rejected , thus x = 7
a 5, b 7, c 8.
I gave up the will to live half way through and got my calculator out, then thought ‘why bother’
I'm right there with ya, my man!
Good job keep going
(√2cos15)^9 is the answer 😅
It is! Can you provide initial setup steps to get trig in there? You are good.
Olympiad Mathematics: 2^(x² - 4x + 2) = 5/4; x = ? 4[2^(x² - 4x + 2)] = 5, (2²)[2^(x² - 4x + 2)] = 2^(x² - 4x + 2 + 2) = 5 2^(x² - 4x + 4) = 2^(x - 2)² = 5, (x - 2)² = log₂5; x = 2 ± √log₂5 x = 2 + √log₂5 = 2 + 1.524 = 3.524 or x = 2 - √log₂5 = 2 - 1.524 = 0.476 Answer check: x = 3.524: x² - 4x + 2 = 4.524² - 4(3.524) + 2 = 0.323 2^(x² - 4x + 2) = 2^0.323 = 1.251 = 5/4; Confirmed x = 0.476: (0.476)² - 4(0.476) + 2 = 0.323; Confirmed The calculation was achieved on a smartphone with a standard calculator app Final answer: x = 2 + √log₂5 = 3.524 or x = 2 - √log₂5 = 0.476
Nice sir
a =5,b=7,c=8
8^a = (2^3)^a = 2^3a √a = 3a a = 9a² 9a² - a = a(9a - 1) = 0 a = 0 , 1/9
Nice it would be great to do a live stream where people can prove math
An approximation based on two terms of a Taylor's series. The more precise result is 0.9935/36,. Nice. Depends on precision you desire.
ab/bc=35/56, a/c=5/8, a=c*5/8, a*a*5/8=40, a*a=25, a=±5.
Their is a rather short way to solve
Let x = 4ᵃ So x³ = (4ᵃ)³ = 4³ᵃ = 64 = 4³ So 3a = 3, i.e. a = 1. So x = 4¹ = 4.
7^5+7^3+7^4+7^2+7=7^3(7^2+1)+7^2(7^2+1)+7=343*50+49*50+7=392*50+7=19600+7=19607
√[64+16√15] = √[a²+2ab+b²] = √[a+b]² = a+b a²+b² = 64 2ab = 16√15 ab = 8√15 = √960 = √[2*2*2*2*2*2*3*5] a=√40=2√10 b=√24=2√6 √[16/(4-√15)] = 2√10 + 2√6
Is it necessary to show a^m+a^n =a^m+n. You think we are stupid and you are the clever one. 4Q
Check the video again. You are wrong.
a 5 b 7 c 8
ABC 578
ab=35.......eqn1 bc=56........eqn2 ca=40.........eqn3 eqn1÷eqn2, we shall get ab/bc=35/56 a/c=5/8.....eqn4 Again eqn3÷eqn2, we shall get ca/bc=40/56 a/b=5/7......eqn5 eqn4×eqn5, we shall get (a/c)×(a/b)=(5/8)×(5/7) a^2/bc =25/56 a^2/56=25/56 (bc=56) a^2=25 a=5 Put the value of a in eqn1 5b=35 b=7 Put the value of b in eqn2 7c=56 C=8 Hence a=5, b=7,c=8
I agree with the comments below that this problem could be easily solved by inspection. However if the numbers were larger and the factors not so obvious there is another interesting way about going about solving this problem Set up a 3x4 matrix [1,1,0,ln(35),0,1,1,ln(56),1,0,1,ln(40)] Do Gauss Jordan Elimination and obtain ln(5), ln(7), ln(8). Raise e to these values. Also realized it it would apply to negative numbers as well. Overkill for this particular problem but it is an interesting approach to these types of problems which at first glance would not seem applicable to linear algebra methods.
You said "without calculator", so I've made : 7x7 = 49 -> sqrt(48) ~= 6.9 sqrt(48) + sqrt(16) ~= 6.9 + 4 = 10.9 ~= 11 sqrt(48) - sqrt(16) ~= 6.9 - 4 = 2.9 ~= 3 11/3 ~= 3.7 it's pretty near from sqrt(3) + 2 ~= 3.73205080757
A mathematical work well done!
:-)
4 ! WTF ?
4