My solution is way easier (it took less than a minute): 1) I rewrote the equation as 1/x+sqrt(1/x+1)=5. 2) Then, I added 1 to both sides of the equation 1/x+1+sqrt(1/x+1)=6. 3) Then, by simply substituting t=sqrt(1/x+1), we get a simple quadratic equation of t^2+t-6=0. The solution of which are -3 and 2. 4) When we plug in the value of sqrt(1/x+1)=-3 has no real solutions ( by the way, if it did not violate the laws of mathematics, the answer choice of 1/8 would be correct). 5) sqrt(1/x+1)=2 1/x+1=4 x=1/3
X=1/8, or 1/3 ^=read as to the power *=read as square root As per the question the 2nd term is *{(x+1)/x} =*{(x/x)+(1/x)} =*{1+(1/x)} As per question (1/x)+*{1+(1/x)}=5 Let (1/x)=R R+*(1+R)=5 *(1+R)=5-R Take the square {*(1+R)}^2=(5-R)^2 1+R=R^2+25-10R R^2-10R-R+25-1=0 R^2-11R+24=0 R^2-8R-3R-24=0 R(R-8)-3(R-8)=0 (R-8(R-3)=0 R-8=0 or R-3=0 R=8 or R=3 1/x=8 or 1/x=3 X=1/8 or x=1/3
1 + √{ x(x+1) } = 5x x(x+1) > 0 ( because x can't be zero and x = - 1 does not satisfy the equation) either x < - 1 or x > 0 also 5x > 1 finally x > 1/5 now x(x+1) = 25 x^2 - 10 x + 1 24 x^2 - 11 x + 1 = 0 (8 x-1) (3x-1) = 0 x = 1/8 , 1/3 but x > 1/5 hence only solution is x = 1/3 we can verify it.
No debemos olvidar que muchas veces por el hecho, que una ecuación nos conduzca a una ecuación de segundo algunas de ellas son soluciones extrañas y haya que comprobar en la ecuación original
There is a term called extraneous roots, in most cases of even roots being square root, 4th root, 6 root and so etc can tend to end up into a result or two that is rejected.... Take for example: x = 4 Square both sides: x^2 = 16 sqr(x^2) = sqr16 x = +/- 4, now (-4) is in this reasoning fact is rejected, because x = 4... Now let us cube both sides of x = 4... cube root is an odd root. x^3 = 4^3 x^3 = 64 cube-root(x^3) = cubeR(64) x = 4, notice we don't run into any issue when use a odd root...
@@patk5724It is not necessarily true that there is no issue with odd roots! If one wants to define the cube root (f.i) of complex numbers it is difficult/unnatural to do in such a way that (-1)^(1/3)=-1. Most mathematical libraries, including Wolfram Alpha, will return a complex number, (-1)^(1/3)=(1+i√3)/2 or a numerical approximation of this number.
Lo hice asignando t=✓(1+1/x) lo que conduce a la ecuación t^2+t-6=0 con t1=-3 y t2=2 se descarta el valor -3 dando como única solución t2=2, y regresando a la variable original en x, y reemplazando t=2 nos queda la ecuación ✓(1/x+1) =2 dónde nos da que x=1/3 es la única solución 😊
My solution is way easier (it took less than a minute): 1) I rewrote the equation as 1/x+sqrt(1/x+1)=5. 2) Then, I added 1 to both sides of the equation 1/x+1+sqrt(1/x+1)=6. 3) Then, by simply substituting t=sqrt(1/x+1), we get a simple quadratic equation of t^2+t-6=0. The solution of which are -3 and 2. 4) When we plug in the value of sqrt(1/x+1)=-3 has no real solutions ( by the way, if it did not violate the laws of mathematics, the answer choice of 1/8 would be correct). 5) sqrt(1/x+1)=2 1/x+1=4 x=1/3
That's a great simplification! 👍Thanks for sharing your clever solution! 🤩
1/x=y
y+rt(1+y)=5
rt(1+y)=5-y
y
X=1/8, or 1/3
^=read as to the power
*=read as square root
As per the question the 2nd term is
*{(x+1)/x}
=*{(x/x)+(1/x)}
=*{1+(1/x)}
As per question
(1/x)+*{1+(1/x)}=5
Let (1/x)=R
R+*(1+R)=5
*(1+R)=5-R
Take the square
{*(1+R)}^2=(5-R)^2
1+R=R^2+25-10R
R^2-10R-R+25-1=0
R^2-11R+24=0
R^2-8R-3R-24=0
R(R-8)-3(R-8)=0
(R-8(R-3)=0
R-8=0 or R-3=0
R=8 or R=3
1/x=8 or 1/x=3
X=1/8 or x=1/3
1 + √{ x(x+1) } = 5x
x(x+1) > 0 ( because x can't be zero and x = - 1 does not satisfy the equation)
either x < - 1 or x > 0
also 5x > 1
finally x > 1/5
now x(x+1) = 25 x^2 - 10 x + 1
24 x^2 - 11 x + 1 = 0
(8 x-1) (3x-1) = 0
x = 1/8 , 1/3
but x > 1/5
hence only solution is x = 1/3
we can verify it.
that argument x>1/5 is crucially missing in the video
Thanks for sharing your perspective 🙏💕🥰
t=1/x...t+√(1+t)=5...1+t=25+t^2-10t...t^2-11t+24=0...t=(11+5)/2=8...t=(11-5)/2=3..x=1/8,1/3..1/8ko
How is it possible that sometimes a solution has to be rejected (in this case x=1/8)? You followed all the right steps...
No debemos olvidar que muchas veces por el hecho, que una ecuación nos conduzca a una ecuación de segundo algunas de ellas son soluciones extrañas y haya que comprobar en la ecuación original
There is a term called extraneous roots, in most cases of even roots being square root, 4th root, 6 root and so etc can tend to end up into a result or two that is rejected....
Take for example:
x = 4
Square both sides:
x^2 = 16
sqr(x^2) = sqr16
x = +/- 4, now (-4) is in
this reasoning fact is rejected, because x = 4...
Now let us cube both sides of x = 4... cube root is an odd root.
x^3 = 4^3
x^3 = 64
cube-root(x^3) = cubeR(64)
x = 4, notice we don't run into any issue when use a
odd root...
@@patk5724It is not necessarily true that there is no issue with odd roots! If one wants to define the cube root (f.i) of complex numbers it is difficult/unnatural to do in such a way that (-1)^(1/3)=-1. Most mathematical libraries, including Wolfram Alpha, will return a complex number, (-1)^(1/3)=(1+i√3)/2 or a numerical approximation of this number.
Lo hice asignando t=✓(1+1/x) lo que conduce a la ecuación t^2+t-6=0 con t1=-3 y t2=2 se descarta el valor -3 dando como única solución t2=2, y regresando a la variable original en x, y reemplazando t=2 nos queda la ecuación ✓(1/x+1) =2 dónde nos da que x=1/3 es la única solución 😊
Thanks for sharing your method 😎💯💕🥰✅💪
Square root of 9 =minus -3 and +3 isnt it? So if we take X=equal to minus 3 ,it verifies the equation
Harvard University Admission Interview Tricks: 1/x + √[(x + 1)/x] = 5; x =?
1 > x > 0; 1/x + √[(x + 1)/x] = 1/x + 1 + √[(x + 1)/x] = 5 + 1
(x + 1)/x + √[(x + 1)/x] = 6, Let: y = √[(x + 1)/x]; y > 0
(x + 1)/x + √[(x + 1)/x] = y² + y = 6, y² + y - 6 (y - 2)(y + 3) = 0, y > 0, y + 3 > 0
y - 2 = 0, y = 2 = √[(x + 1)/x], (x + 1)/x = 2² = 4, 4x = x + 1; x = 1/3
Answer check:
x = 1/3, 1/x = 3: 1/x + √[(x + 1)/x] = 3 + √[3(1/3 + 1)] = 3 + 2 = 5; Confirmed
Final answer:
x = 1/3
1/X + [(X+2)/X]^1/2 = 5
[(X+1)/X]^1/2 = 5 - 1/X
[[(X+1)/X]^1/2]^2 = (5 - 1/X)^2
(X+1)/X = [(5X -1)/X]^2
(X+1)/X=(25X^2 -2×5X×1+1^2)/X^2
Умножим обе стороны уравнения на X^2
X(X+1)=25X^2 - 10X + 1
X^2 +X = 25X^2 -10X + 1
25X^2- 10X + 1 - X^2 -X = 0
24X^2 - 11X + 1 = 0
D= (-11)^2 - 4×24×1= 25
D^1/2= 5
X=(11+5)/(24×2)=1/3
X=(11-5)/48=1/8
Ответ: 1/3 и 1/8
D
Sir you are very very very slow
Thanks for the feedback! I'm working on making the videos more concise.I'll try to explain the concepts more quickly in future videos.🙏