Great 👍

(x ➖ 3x+2)

Nice

6sa

Sardar20024

(x ➖ 3x+2)

Nice ❤

Cube root means vectors addition class. You can do geometric construction to do that.

I just substituted numbers in my head. When I got to 2 and 4 it all fit…

I always taught mathletes to try -1,0,1 immediately. This saves a lot of unnecessary time. If you solve a problem without explaining why you do something, it doesn't help novice problem solvers. Here one term is positive and the other is negative. What are the implications?

Far too complicated solution. 1) you can do it in your head. What two squares add up to 20? 16 and 4. 2) a=6-b (6-b)^2+b^2 = 20 36-12b+b^2+b^2 = 20 2b^2-12b+16 = 0 b= (12+/- sqrt(12^2-4*2*16))/2*2 b=(12+/- sqrt(16))/4 b=(12+/-4))/4 b=4 or 2 and since a=6-b it follows a= 2 or 4.

If x is a variable,dont use x for multiplication. Use a dot!!

The solution is fine. But if this a teaching moment, the plan is the most important part of the solution!!

4/ab^4 =1 (b ➖ 1a+1)

(x ➖ 3x+3)

My answer is 1+or- 5√5

Great we got 4 solutions for x. Now let's plug them back into the equation to see if they make the equation equal 0. All the solutions are too hard for me to plug them back in, will somebody please check them for me?(oh kudos to me for being the 1st one to leave a comment on here too!)

(x ➖ 5x+2)

(x ➖ 3x+2) (x ➖ 3x+2)

(x ➖ 2x+2)

bcos○+bcos○➖ =bcos○^2Logasin○+asin○➖ =asin○^2 {bcos○^2+asin○^2}= bcos○asin○^4 bcos○asin○^2^2 (bcos○ ➖ 2asin○+2) (bcos○)^2=bcos○^2 (asin○)^2=asin○^2 {bcos○^2 ➖ asin○^2)= bcos○^0asin○^0 bcos○^1asin○^1 (bcos○ ➖ 1asin○+1)

7/ab (b ➖ 7a+7)

(x ➖ 2x+2)

Thanks sir, g❤❤❤

Great video❤❤❤

Great video❤❤❤

Great video❤❤❤

Great video❤❤❤

Great video❤❤❤

The background noise is annoying

your square roots are dreadful. they look like 5 a square root symbol is formed from 3 straight lines

Nice question sir

nice question which standard question

Great question and interesting sir, g❤❤❤

Interesting sir,g❤❤❤

It's simpler to put it like this... The terms are grouped around (x+9)^4 so substitute t=x+9. Note that the +- 2 and +-1 cancel the odd terms so (t-2)^4+(t+2)^4 + (t-1)^4+(t+1)^4 + t^4=99 2*t^4+48*t^2+32 + 2*t^4+12*t^2+2 +t^4 = 99 simplifying ... 5t^4+60t^2-65=0 or (dividing by 5) t^4+12t^2+13=0 (t^2-1)(t^2+13)=0 t (=x+9) = ± 1 or ± i√13 so subtract 9 from the list to get x

Interesting❤❤❤

Muy interesante video, muchas gracias por compartir. Muy buena y didáctica explicación, mi hija y mi persona estamos muy agradecidos. 😊😊

I liked your method but brute force and ignorance also works. Multiply both sides by (1+√x)^4 and expand using binomial (coeffs 1,4,6,4,1) => 82*√x^4-120*√x^3+108*√x^2-120*√x+82=16*√x^4+64*√x^3+96*√x^2+64*√x+16 66*√x^4-184*√x^3+12*√x^2-184*√x+66=0 33*x^2-92*x√x+6*x-92*√x+33=0 This is symmetrical so divide by x and group terms by x and √x 33(x+1/x)-92(√x+1/√x)+6=0 if we put s=(√x+1/√x) then s²=x+1/x + 2 => 33(s²-2)-92s+6=0 33s²-92s-60=0 (3s-10)(11s+6)=0 s=10/3 or -6/11 (but -6/11 gives imaginery roots) s=10/3=(√x+1/√x) multiply by 3√x 10√x=3x+3 3x-10√x+3=0 (3√x-1)(√x-3)=0 √x=1/3 or 3 x=1/9 or 9

Let x=sqrt(a) then sqrt(-a) = ix equation becomes x + ix = 32 x(1+i) = 32 x = 32/(1+i) multiply top and bottom by (1-i) x = 32(1-i)/(1+i)(1-i) x = 32(1-i)/2 = 16(1-i) square both sides a = x^2 = 256(1 - 2i - 1) = -512i

Interesting question ❤❤❤❤

we mutiply the numerator and denominator by √x+2 we will have √((√x+2)²/(x+4)) = √x+2 so √(√x+2)²/√(x-4) = √x+2 and then √x+2/√(x-4) = √x+2 we simplify 1/√(x-4) = 1 ==> √(x-4) = 1 x-4 = 1 ==> x= 5

let sqrt(sqrt(x)+11) = t, then sqrt(x) = t^-11, then t+sqrt(t^2-11) = 11, then sqrt(t^2-11) = 11-t, then t^2-11=11^2-22t+t^2, then t=6. Then sqrt(x)=25 then x = 625

Nice, but I think this way is simpler here. Let _a = √x + 2, b = √x - 2_ _a ≥ 2_ and _ab = x - 4_ The equation is _√(a/b) = a_ Squaring: _a/b = a²_ ⇒ _a(ab - 1) = 0_ ⇒ _a = 0 (rejected, a ≥ 2) or ab = 1_ ∴ _ab = x - 4 = 1_ ⇒ *_x = 5_*

Interesting question❤❤

V nice sir

🤪👍👍😁🤪👋

Nice

Interesting question ❤❤❤

The second half is unnecessary on this problem. We know a + b = -2 ab = -2 so by Viete we can solve for a and b as a = 1 + sqrt(3), b = 1 - sqrt(3) Then the binomial theorem allows us to show that x = 1 + 3 * 21 + 9 * 35 + 27 * 7 = 1 + 63 + 315 + 189 = 568