x + √x = 1 We want a 1/√x, so divide by √x. √x + 1 = 1/√x Now, add x to both sides, so the RHS is the desired expression x + √x + 1 = x + 1/√x Note that x + √x = 1 1 + 1 = x + 1/√x 2 = x + 1/√x
substitution for √x = y also works, because then we get a quadratic and solve for y = (1 + √5)/2 (second option is rejected because y > 0) , then just insert it to the second equation and evaluate: y^2 + 1/y = (6 - 2√5)/4 + 2/(√5 - 1) = (6 - 2√5)/4 + 2(√5 + 1)/4 = 2
@@jamesharmon4994 Nope. The square root of a positive number is defined as a positive number. This prevents the square root function from being multivalued, which makes life easier.
@@ianchristian7949 your comment makes sense. Square root has two solutions one of them less than zero. However, working with the negative one satisfies the two equations and gives the same result. I also think that square root of x doesn't have to be greater than 0.
Alot simpler than becoming complicated: With knowing x + √x = 1 then x = 1 - √x is the substitution to be used in x + 1/√x or x + 1/(1 - x) = 1 ... Common denominator work on the left side has 1 = (x^2 - x - 1)/(x - 1) ... Cross multiply and x^2 - x - 1 = x - 1 ... x^2 = x ... x = +/- 2 but x= -2 has no real square root and x = 2 is the final answer!
x + √x = 1 We know x > 0, so we divide by √x to get: √x + 1 = 1/√x Now we can compute this: x + 1/√x = ? = x + (√x + 1) = (x + √x) + 1 = (1) + 1 = 2 Alternatively, we see the equation is quadratic in √x, so we solve with the quadratic formula: x + √x = 1 √x = (-1±√5)/2 We keep √x > 0: √x = (√5-1)/2 We compute its square and reciprocal: x = (3-√5)/2 1/√x = (√5+1)/2 Their sum is: x + 1/√x = (3-√5 + √5+1)/2 = 2 We're having so much fun with Φ & φ.
x + √x = 1
We want a 1/√x, so divide by √x.
√x + 1 = 1/√x
Now, add x to both sides, so the RHS is the desired expression
x + √x + 1 = x + 1/√x
Note that x + √x = 1
1 + 1 = x + 1/√x
2 = x + 1/√x
x+1/sqrt(x) = x+(x+sqrt(x))/sqrt(x) = x+sqrt(x)+1 = 1+1 = 2. So simple.
substitution for √x = y also works, because then we get a quadratic and solve for y = (1 + √5)/2 (second option is rejected because y > 0) , then just insert it to the second equation and evaluate:
y^2 + 1/y = (6 - 2√5)/4 + 2/(√5 - 1) = (6 - 2√5)/4 + 2(√5 + 1)/4 = 2
Useful to remember that phi^2 = phi+1. Saves a lot of √5 arithmetic.
Good point!
@SyberMath: At 03:45, you used the wrong sign in the rightmost denominator. You have √5-1 instead of √5+1 .
At 3:45 when you rationalized, it should be root 5 + 1...tiny error. Love your videos bro, thank you!
@@yogesh193001 np. Thank you!
I like your videos very much :)
@@Maths_3.1415 glad to hear that
@@SyberMath
😮
If sqrt(x) = u, then u CAN be negative since the square root of a positive CAN be negative
For example... x + sqrt(x) = 0 if x = 1 and you take the negative square root of 1.
@@jamesharmon4994 Nope. The square root of a positive number is defined as a positive number. This prevents the square root function from being multivalued, which makes life easier.
I solved it with the second method ❤❤
Why does the square root of x have to be greater than 0?
The real sqrt(x) needs to be positive by definition. It’s single-valued unlike the complex roots
@@SyberMath But e.g. -2 is a real square root of 4 but it's
@@ianchristian7949 sqrt is a function that gives the absolute value
@@ianchristian7949 your comment makes sense. Square root has two solutions one of them less than zero. However, working with the negative one satisfies the two equations and gives the same result. I also think that square root of x doesn't have to be greater than 0.
@@tonybluefor this is exactly why the quadratic formula has plus/minus the square root - the "minus" accounts for the negative square root.
Alot simpler than becoming complicated: With knowing x + √x = 1 then x = 1 - √x is the substitution to be used in x + 1/√x or x + 1/(1 - x) = 1 ... Common denominator work on the left side has 1 = (x^2 - x - 1)/(x - 1) ... Cross multiply and x^2 - x - 1 = x - 1 ... x^2 = x ... x = +/- 2 but x= -2 has no real square root and x = 2 is the final answer!
x + √x = 1
We know x > 0, so we divide by √x to get:
√x + 1 = 1/√x
Now we can compute this:
x + 1/√x = ?
= x + (√x + 1)
= (x + √x) + 1
= (1) + 1
= 2
Alternatively, we see the equation is quadratic in √x, so we solve with the quadratic formula:
x + √x = 1
√x = (-1±√5)/2
We keep √x > 0:
√x = (√5-1)/2
We compute its square and reciprocal:
x = (3-√5)/2
1/√x = (√5+1)/2
Their sum is:
x + 1/√x
= (3-√5 + √5+1)/2
= 2
We're having so much fun with Φ & φ.
(A) x + sqrt x = 1
then sqrt x = 1-x
Substitute in (B)
x + 1÷(1-x) = ... = 2
Brži metod je iz jednakosti izraziti koliko je koren iz x i uvrstiti u izraz čiju vrednost računamo.Vrlo brzo se dobije vrednost 2
Just divide the whole first equality by sqrt(x) and you get an expression for 1/sqrt(x) just substitute and you get 2.
this video was funny
Evaluating An Algebraic Expression: x + √x = 1; x + 1/√x = ?
First method:
1/√x = (x + √x)/√x = √x + 1, x + 1/√x = x + √x + 1 = 1 + 1 = 2
Second method:
x + √x - 1 = 0, √x = (- 1 ± √5)/2 > 0; x = (3 -/+ √5)/2
x + 1/√x = (3 -/+ √5)/2 + 2/(- 1 ± √5)] = (3 -/+ √5)/2 + (1 ± √5)/2 = 2
2, some of the methods I’m seeing in the comments are really overly fancy
√x = 1 - x
x = x² - 2x + 1
x² - 3x + 1 = 0 => x² = 3x - 1
x + 1/√x = A = x + 1(1 - x) = x - 1(x - 1)
A = (x² - x - 1)/(x - 1)
A = (3x - 1 - x - 1)(x - 1)
A = (2x - 2)/(x - 1)
*A = 2*
Please stop saying 2b or not 2b, 2c or not 2c ..... This is very childish and not funny thanks
Do u not c what I c? 😁