at 3:47 you break 84 into 64+16+4. that is the key point of the method. but how did you choose those numbers. My method is to look for the largest integer square less than 84 which is 4^3=64 so then with x=4 we find x²+x= 16 +4=84-64. The hard part of maths is not the manipulation it is the direction.
X=2^60......May be ^=read as to the power *=read as square root As per question X^(1/10)+x^(1/15)+x^(1/30)=84 So, X^(6/60)+x^(4/60)+x^(2/30)=84 So, {(X^(1/60)^6+{x^(1/60)}^4+{x^(1/60)}^2=84 Let x^(1/60)=R So, R^6+R^4+R^2=84 Though the equation is sixth powered, so H&T method is applicable If R=2, then (2^6)+(2^4)+(2^2) =64+16+4=84, satisfies the equation Hence R=2 So, X^(1/60)=2 {X^(1/60)}^60=2^60 X=2^60......((Ans)
²³ In the three - (x)^(1/10), (x)^(1/15), (x)^(1/30), the last one is the least index-numbered. Call a = x^(1/30). Then x^(1/10) = [x^(1/30)]³ = a³; x^(1/15) = [x^(1/30)]² = a². ∴ a³+a²+a¹ = 84 which can be expressed as sum of cube, square & itself is: 84=4³+4²+4¹=64+16+4. Then, (a³ -4³)+(a² -4²)+(a -4) = 0. Now reduction of the factors as lesser powered(indexed) ones. Taking (a -4) as common, (a -4)[(a²+4a+16)+(a+4)+1] =0. (a -4)[(a²+5a+21)] =0. One solution is (a -4) =0 → a = 4. Now the quadratic: a²+5a+21 =0; in which discriminant D =√[(-5)² -4 . 1 . 21] =√[25 -84] = √(-59) which is not real. solutions: ½[-5±√-59] - not real solutions. i)a = x^(1/30)= 4 [x^(1/30)]³⁰=4³⁰ = x. x = (2²)³⁰= 2⁶⁰ =(2¹⁰)⁶ = 1024⁶ = 1.152925505 X 10¹⁸.
at 3:47 you break 84 into 64+16+4. that is the key point of the method. but how did you choose those numbers.
My method is to look for the largest integer square less than 84 which is 4^3=64 so then with x=4 we find
x²+x= 16 +4=84-64.
The hard part of maths is not the manipulation it is the direction.
Your method is great! ❤
X=2^60......May be
^=read as to the power
*=read as square root
As per question
X^(1/10)+x^(1/15)+x^(1/30)=84
So,
X^(6/60)+x^(4/60)+x^(2/30)=84
So,
{(X^(1/60)^6+{x^(1/60)}^4+{x^(1/60)}^2=84
Let x^(1/60)=R
So,
R^6+R^4+R^2=84
Though the equation is sixth powered, so H&T method is applicable
If R=2, then
(2^6)+(2^4)+(2^2)
=64+16+4=84, satisfies the equation
Hence R=2
So,
X^(1/60)=2
{X^(1/60)}^60=2^60
X=2^60......((Ans)
Very nice! ❤
²³
In the three - (x)^(1/10), (x)^(1/15), (x)^(1/30), the last one is the least index-numbered.
Call a = x^(1/30). Then x^(1/10) = [x^(1/30)]³ = a³; x^(1/15) = [x^(1/30)]² = a².
∴ a³+a²+a¹ = 84 which can be expressed as sum of cube, square & itself is: 84=4³+4²+4¹=64+16+4.
Then, (a³ -4³)+(a² -4²)+(a -4) = 0. Now reduction of the factors as lesser powered(indexed) ones. Taking (a -4) as common,
(a -4)[(a²+4a+16)+(a+4)+1] =0.
(a -4)[(a²+5a+21)] =0.
One solution is (a -4) =0 → a = 4.
Now the quadratic: a²+5a+21 =0; in which discriminant D =√[(-5)² -4 . 1 . 21] =√[25 -84] = √(-59) which is not real.
solutions: ½[-5±√-59] - not real solutions.
i)a = x^(1/30)= 4
[x^(1/30)]³⁰=4³⁰ = x.
x = (2²)³⁰= 2⁶⁰ =(2¹⁰)⁶ = 1024⁶ = 1.152925505 X 10¹⁸.
Very nice! ❤