Відео

Okay

3^(×-1)=5^(1-×^2) =>3=5^(1+×) =>×={(log @5(3)-1}

I don't want to be cocky but this peoblem us way too easy,for Olympiad

Quite a easy one

Too much talking in very easy places

We can solve it, so the answer is likely to be nice. Root 15~4, so approx 8^x+0=62. Thus x is about 2. Check 2. The middle trrms cancel, giving 16+15+15+15=62.

5³ᵃ = 10 3a · lg(5) = 1 a = 1 / [3 · lg(5)] a ≈ 0,477 (which is very close to lg(3), by coincidence)

😂(×+y)(×-y)=98/2=49 49=49(1)or 7(7)(n.a.) ×+y=49&×-y=1 =>×=25&y=24

Isn't the answer 1.

a=ln(10)/(3ln(5)). This is pretty easy to generalize.

{(x^3)^2-[(x-1)^3]^2}=0 [x^3-(x-1)^3][x^3+(x-1)^3]=0 {(1)[x^2+x(x-1)+(x-1)^2]}{(2x-1)[x^2-x(x-1)+(x-1)^2]}=0

62=16+15+16+15=16+15+8*15+16+15-8*15=4^2+(*15)^2+2×4×*15+4^2+(*15)^2-2×4×*15=(4+*15)^2+(4-*15)^2 (4+*15)^x+(4-*15)^x=(4+*15)^2+(4-*15)^2 (4+*15)^x=(4+*15)^2buradan x=2 (4-*15)^x=(4-*15)^2 buradan da x=2. Təşəkkürlər.

The equation reduces down to (x² + 1 + 1/x²) / (x + 1 + 1/x) = 3 Let u = x + 1 + 1/x u² = (x + 1 + 1/x)² = (x² + 1 + 1/x²) + 2(x + 1 + 1/x) = (x² + 1 + 1/x²) + 2u u² - 2u = (x² + 1 + 1/x²) The equation (x² + 1 + 1/x²) / (x + 1 + 1/x) = 3 becomes (u² - 2u) / u = 3 u = 5 x + 1 + 1/x = 5 x² - 4x + 1 = 0 x = 2 ± √3

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Approximation method,x=2 insertion can be easy.But if the question x=more than 10,you need more difficult tackling.

12:23

(4 + roo15) ² + (4-root15)²=2(16+15)=62 so x=2

Why is x-3=0?

Verrrry verrrty good tank you🙏🙏🙏

you can totally recognise the pattern which is 5! = 120 then the value of x is undeniably 1, so easy

Alternatively, the second integer solution of x=-6 can also be easily "seen" from the original form of the equation in much the same way that we spotted x=1. If you write the 120 = 5! = (-5)*(-4)*(-3)*(-2) = (x+1)(x+2)(x+3)(x+4), then you will immediately recognize that x+1 = -5 will satisfy it. Thus x=-6 is a solution.

It's a quartic equation with real coefficients (x^4 + 10x^3 + 35x^2 + 50x - 96 = 0). It has two integer solutions and two complex-conjugate solutions. The two integer solutions can be easily guessed. The first one (x=1) can be guessed from the original form of the equation, the second solution (x=-6) can be guessed after splitting off the linear factor (x - 1) and testing the integer divisors of the last coefficient of the resulting cubic equation (x^3 + 11x^2 + 46x + 96 = 0). After splitting off another linear factor (x + 6), we arrive at a quadratic equation (x^2 + 5x + 16 = 0). That one can be solved by the usual formula. It doesn't have any real solutions but we get the two complex-conjugate solutions. They both have a real part of -5/2 and an imaginary part with an absolute value of 1/2*sqrt(39). Easy.

1 and -6 are easy solutions to get by asking what four numbers in a sequence increasing by 1 are the same as 2 * 3 * 4 * 5. you can see 1 will give you this sequence from this problem. And notice that if all the numbers are negative you get the same solution so -5 * -4 * -3 * -2 this sequence can be found by putting in -6. The other 2 solutions appear to be imaginary or complex from me quickly looking at it. We know that the angles of the 4 consequent numbers must be added together to make the overall angle 0. the issue is the angle will change between the four numbers. Meaning no more eyeballing the problem

I just looked at the prime factorization of 120 (2^3*3*5) and realized u can use this for x=1(2*3*4*5) or - 6 (u can put a - sign on each of 2,3,4,5 bc they r an even number and it cancels out). Then I multiplied the given expression out (x^4+10x^3+35x^2+50x+24-120=0) and did a polynomial long division with the one given by the 2 found solutions ( (x-1)*(x+6) = x^2+5x-6), which gives x^2+5x+16. With the quadr. form. we get (- 5 +/- sq(39)i)/2 for the other 2 solutions.

1? 120 can be decomposed into a multiplication of primes. : 2 × 2 × 2 × 3 × 5 .. so it was a guess.

Good one

this equation was resolved by substitution from a 4th degree power to quadratic equation via k, is there a way we can do this for cubic equation?

Very intelligent

If we add both entities like m raised to 2 -n and later term we can get m raised 2- m + n raised 2-n=98 By hit and trial we can satisfied above equation with takin value 7 and 8 for m and n

useful formula x^3 + y^3 = (x + y)^3 - 3xy(x + y) ax^2 + bx + c = 0 and b is even(b = 2b') => x = (-b' ± √(b'^2 - ac))/a

My approach is: (a+b)(a-b)=4 (a^2+b^2+4)(a^2+b^2-4)=16 (a^2+b^2)^2=32 a^2+b^2=4root(2) (notice that a^2 and b^2 are both positive, so no plus or minus sign is needed for this square root) (a+b)^2=4root(2)+4 a+b=plus or minus 2root((root(2)+1))

i got a+b = root( root(32)+4 ), is it correct or where did i do it wrong

I am bored so lets solve with complex numbers. 4+4i=a²+2abi-b²=(a+bi)² a+bi=√(4+4i) Let z=4+4i |z|=4√2 z/|z|= 1/√2 + i/√2 Arg(z)=π/4 Therefore √z=2∜2 * e^(πi/8) Which as shown earlier is equal to a+bi. a=Re(√z), b=Im(√z) a=2∜2 * sin(π/8) b=2∜2 * cos(π/8) a+b=2∜2 * √2 sin(⅜π) a+b= 2^(¾) * √(2+√2) а+b=2√(1+√2) Naturally while taking the square root of z instead of Arg(√z)=⅛π we can use Arg(√z)=⅝π which will by similar methods turn out to be the other solution in which a+b=-2√(1+√2)

A Very Nice Math Olympiad Problem: x³ + x² = 4/27; x =? x³ + x² - 4/27 = 0; (x³ - 1/27) + (x² - 3/27) = [x³ - (1/3)³] + [x² - (1/3)²] = 0 Let: a = 1/3; [x³ - (1/3)³] + [x² - (1/3)²] = (x³ - a³) + (x² - a²) = 0 (x - a)(x² + ax + a²) + (x - a)(x + a) = (x - a)(x² + ax + a² + x + a) = 0 x - a = 0, x = a or x² + ax + a² + x + a = x² + 4x/3 + 4/9 = (x + 2/3)² = 0, x + 2/3 = 0 x = 1/3 or x = - 2/3; Double root Answer check: x = 1/3: x³ + x² = 1/27 + 1/9 = 4/27; Confirmed x = - 2/3: - 8/27 + 4/9 = 4/27; Confirmed Final answer: x = 1/3 or x = - 2/3; Double root

√15≈√16 so: 4−√15≈0 and 4+√15≈8 0²+8²=64 and 64≈62 Let's try x=2 (4+√15)²+(4−√15)²=(16+8√15+15)+(16−8√15+15) (4+√15)²+(4−√15)²=32+30=62 x=2 is indeed a solution of f(x)=(4+√15)ˣ+(4−√15)ˣ−62 f(x)=nˣ+1/nˣ−k is symmetrical with respect to the origin and has 2 solutions as a sum of two exponentials, one increasing and the other decreasing. If x is a solution of f(x)=nˣ+1/nˣ−k then -x is the other solution of f(x)=nˣ+1/nˣ−k Here we have n=4+√15 and 1/n=4−√15 because (4+√15)(4−√15)=4²−(√15)²=1 f(x)=nˣ+1/nˣ−62 having solution x=2 so x=-2 is also a solution.

Where did you come up with -2 as a root of the cubic, though, lucky guess?

👍👍

I love how you did both a log version and a pure exponent version. The exponent version required the 6 to show up where it did, tho. You couldn't use that method if there were different numbers/ bases on each side

I don’t see how you concluded a-2 and a-3 were factors except by intuiting that 97 is the sum of two numbers which are 2 perfect quartic roots. And since you can restate the problem as: find all x,y such that x,y are natural numbers, x+y=97, and x^.25 + y^.25 = 5, then brute force is faster than the binomial expansion, 2 polynomial divisions, and one quadratic solve. Even when replacing 97 with 195857 and 5 with 34, brute force will be faster.

Factor when ever you can. Good idea to start solving.

Love the math you showed. I solved it just be realizing that they have to be integers and you need to end on an integer so they have to be perfect squares as well as positive (or you'd have an i on the RHS), so just cycle through them starting with 1, 4, 9, and there's your answer: 4, 9

10:28 I would have reduced the 2 from the bottom before I split into 2 cases.

Really excellent, although to be fair the original was contrived to be able to simplify this far. Still loved your work. I think I would have gone from (root 9)^2 => 9 a little quicker and once I had shown the trick went from (root 9)^2 - (root 5)^2 => 9 - 5 = 4. But that's just a style choice

This one was easy i got the ans under 2 min !!! thank you for the question !

Wonderful job. You could have given the approximate values which are close to -7.5 and +6.5.

Hey can anyone tell me what i am doing wrong here , x^6 = x^4 + x^2 x^6 - x^4 - x^2 = 0 x^3 - x^2 - 1 = 0 x ( x^2 - 1 ) = 1 x cant be 1 as it will form a equilateral triangle and we know angle given is 90 so x^2 - 1 = 1 so x^2 = 2 and x = root2 and x = root2 doesnt satisfy (x^3)^2 = (x^2)^2 + x^2

Interestingly, Desmos thinks there is only one solution to the equation

this question looked so simple on the first glance , great explanation and wonderful question choice

X^3 + X^2 - 4/27 = 0 --- (1) X = 1/3 ( by trial and error ) divide eq (1) by (x - 1/3) get the other value of x as -2/3 and -2/3 ( twice ) so finally three values of x are 1/3 , -2/3 , -2/3 Edit - personally i think this is a better way and quicker way , let me know if im doing something wrong

p^2-62p+1=0 çevrilmiş kvadrat tənlik olduğundan p=62/2 +*(62/2)^2-1=31+*31^2-1=31+*961-1=31+*960=31+*960=31+*64×15=31+8*15=(4+*15)^2; p=31-8*15=(4-*15)^2 (* kökaltı kimi başa düşülsün) kimi həll etsə idiniz kiçik ədədlərlə işləməli olardınız ki, bu da Sizin əməyinizi yüngülləşdirərdi. Təşəkkürlər. (* kökaltı