You should have mentioned that (1 + sqrt(5))/2 = φ, the Golden Ratio. It's a special number such that squaring it has the same result as adding 1 instead. φ^2 = φ+1, and we can use this to swap the x^2 and (x+1) in the original equation: (φ+1) - 1 = sqrt(φ^2) φ = φ
if you set one side of the original eqn to y you would get a nice system of equations x=y^2-1 y=x^2-1. You can then use simple algebra and get (x-y)(x+y+1)=o This avoids having to solve a cubic like it was shown.
Yes, exactly, if you recognize the left hand side as a difference of two squares you can simplify the equation without having to deal with the fourth degree polynomial.
Conclusion from this video: high school in the Netherlands has an atrocious quality. I never learned how to solve such a cubic equation. Adding some arbitrary X² and then finding a common factor to split it up. I see what he is doing, I am not sure about the logic behind it. It seems that the goal is to rewrite the equation in a form which is solvable by hand but how do you figure out how to get there? What if you wouldn't be able to get that common factor?
It's not jst abt this equation or abt this solution or abt the idea of this prob it is abt the logic abt the how you will get the idea? and i think the answer is the habit of doing math problems and there is not only this way to solve it what i want to say is u jst need to have the math logic and think of athor ways bcz most of tge times we don t get it from the frst try so we jst need to move the pen .
That's my question too. We see this manipulation all the time but it's never properly explained. Which is weird because they over-explain everything else but never this.
Cubic equations may be taken up by Trial & Error Method of factorisation in which +ve & -ve values like 1, 2, 3, ... are assigned to satisfy. Of course, for a root like x = -73.89 this method cannot be followed, rather by Sridhar Acharya or by computer to solve.
So the OP is a bit of an idiot. Look at x^3-2x-1. Plug in factors of the linear term - 1 and -1 - and we find (x+1) is a factor. So divide (x+1) and get (x^2-x-1). The +x^2-x^2 trick is just a fancy way to divide by (x+1)
Es mejor pasar todo a un miembro de la ecuación y extraer factor común √(x+1). De hecho , creo que es la manera correcta, por su eficiencia, de resolver esta ecuación: (x+1)(x-1)-√(x+1)=0 √(x+1) [(x-1)√(x+1) - 1]=0 Del primer factor obtenemos x=-1, que comprobamos y resulta ser válida. Del segundo, elevamos al cuadrado en: (x-1)√(x+1)=1 para obtener, x(x²-x-1)=0 cuyas soluciones son: x=0, que descartamos, x=(1+√5)/2 (válida), y x=(1-√5)/2, que también descartamos tras su comprobación.
Before watching the video: One root is -1 (easy), the second is ~ +1.618 (calculated numerically) When I divide the equation by the first root, I get (x-1)*SQRT(x+1)=1 Squaring that I get x³-x²-x=0 => x(x²-x-1)=0 x=0 does not work in the original equation The quadratic equation has two roots 0.5*(1+/- SQRT(5)) The + variant works, the - variant does not. => The roots are -1 and 0.5*(1+SQRT(5)).
That's the way I did it too. It's also interesting that -1 no longer works as a solution to x^3-x^2-x=0 due to the earlier step of dividing by √(x+1), which constitutes division by zero when x=-1.
You can see from the graph that the two rejected "solutions" correspond to where x²-1 and √(x+1) are of opposite value. Which makes sense, as the square of a negative number is the same as the square of its positive counterpart, and our first manipulation of the equation was to square both sides...
Simplify x-1 = 1/(x+1)^0.5. Then square both sides. x^2 -2x + 1 = 1 / (x+1) Transpose (x+1)(x^2 −2x+1)=1. Expand and simplify x^3−x^2−x=0 or x(x^2 −x−1)=0. Roots. (1) x =0. Extraneous. By quadratic equation formula, (2) x = 1.618 acceptable ans. (3) x = - 0.618 acceptable ans.
(In case you don’t understand LaTeX notation, I’ve edited the comment and included ASCII equations next to the LaTeX expressions) This problem is very elementary and *does not* require one to find the roots of a cubic polynomial. The equation has two real roots of which one is obviously $-1$ ( -1 ). To find the positive real root, first, factor the LHS of the equation : $(x + 1)(x - 1) = \sqrt{x + 1}$ ( (x + 1)(x - 1) = sqrt(x + 1) ) Now square both sides : $(x + 1)(x - 1)(x + 1)(x - 1) = (x + 1)$ ( (x + 1)(x - 1)(x + 1)(x - 1) = (x + 1) ) Next, divide both sides by $(x + 1)$ ( (x + 1) ) : $(x + 1)(x - 1)(x - 1) = 1 \ \text{for} \ x e -1$ ( (x + 1)(x - 1)(x - 1) = 1 , for x =/= -1 ) Now, expand the LHS : $(x^{2} - 1)(x - 1) = x^{3} - x - x^{2} + 1 = 1$ ( (x^2 - 1)(x - 1) = x^3 - x - x^2 + 1 = 1 ) Subtract $1$ ( 1 ) from both sides : $x^{3} - x - x^{2} = 0$ ( x^3 - x - x^2 = 0 ) Divide by $x$ ( x ) : $x^{2} - 1 - x = 0$ ( x^2 - 1 - x = 0 ) Solve for $x$ ( x ) : $x = \dfrac{1}{2} + \dfrac{\sqrt{5}}{2}$ ( x = 1/2 + sqrt(5)/2 )
Think of the implied geometry: recognize that what you are looking at is an equation asking for where a function and its inverse are equal. They are equal on the line y = x. The solution is then very simple as you can solve x^2 - 1 = x or sqrt( x + 1 ) = x. My guess is that this is the intended method of solution as it is not tedious.
@@oahuhawaii2141 if you think of the graphs, the -1 solution is obvious and the positive solution follows this rule. You can even easily determine which of the two solutions is rejected. (I was able to solve in my head because of this.)
@@gregorylewis4426: But the graph isn't a solution. Rather, it merely shows where the solutions lay, and they still must be found. The solution OP gave is lacking, and simply saying "it's easily determined" isn't going to cut it on an exam: I didn't solve the problem because it's obvious or easily found.
@@oahuhawaii2141More formally, for negative x values, you have a monotonically decreasing function and a monotonically increasing function, so there can only be one intersection, and we can easily see that the x value for that is -1. For the non-negative half, we have a function and its inverse, so we can set x=y. From there we have x^2-1=x, so x^2-x-1=0, so x=(1+-√5)/2, and since (1-√5)/2 is negative, we reject that, giving us our two solutions. Neither solution is better. This one is just much simpler.
A and B beeing two reals, we have: sqrt(A) = B is equivalent to A = B^2 and B>=0. So the given equation is equivalent to (x^2 -1)^2 = x + 1 and x^2 - 1 >=0 (i.e. x is in E = ]-infinity, -1] U [1, +infinity[ ) We solve (x^2 -1)^2 = x + 1, and among the values of x that we obtain, the solutions of the given equation are the ones beeing in E.
There is a slightly easier way to solve this, where we don't have the problem to solve a cubic equation. Starting with squaring both sides we get (x^2-1)^2=x+1 Now we use the second binomial formula on the left side to factor : (x-1)^2*(x+1)^2=x+1. We see both sides have the factor x+1. We check if x+1=0 is a solution, which is indeed the case. Therefor x1=-1. Case 2: x-1, that means x+10. In this case we can divide both sides of the equation by x+1. We get (x+1)(x-1)^2=1. Now we expand (x+1)(x^2-2x+1)=x^3-2x^2+x+x^2-2x+1=x^3-x^2-x+1=1 Subtracting 1 on both sides gives us x^3-x^2-x=0. We can now easily see that we can factor x: x(x^2-x-1)=0. We check x=0 in the original equation and see it is not a solution. Therefor we can divide both sides by x: x^2-x-1=0 It's the same quadratic equation like in the video which gives us the same solution, the (1-sqrt(5))/2 we reject, because checking in the original equation reveals that this is not a solution. 2 things to note: if you square both sides of an equation you always have to check if your solutions are true solutions because this step can give you extraneous solution, which indeed happened in this case. And second if you divide both sides of an equation by a term containing x, always check, if this term can equal zero. Because dividing both sides of an equation by zero is not allowed and in this case you can lose solutions. Would we have just divided by x+1 without checking, we would have lost the solution -1.
Square both sides (x+1)(x-1)(x-1)(x+1)=(x+1) Set x+1=0, x=-1 Divide both sides by x+1 (X2-1)(X-1)=1=X3-x2-x+1 X3-x2-x=0=x(x2-x-1) X=-1,0,(1+-{5})/2 but 0 and (1-{5})/2 are artifacts from squaring both sides
Good work, but since x = -1 is a trivial solution easily found by plug and check, we could reduce the amount of work by dividing the 4th degree equation down to a cubic right off the bat.
√x² = x is not an identity. Use this substitution in an equation is a heavy error. Use this substitution in an educational video is uneducational.. √x² = |x| is an identity, good for all values of x.
@ So you explain high schooler a *wrong* stuff? A wrong thing they remember forever, causing them to make mistakes in the future? I don’t think so. I think it’s a limit of the author’s teaching abilities. Obviously in this case the step works, the two equations (before and after the substitution) are equivalents, but only for the specific properties of this case. You have to specify this, if you love your students.
When does he says that? If the power is outside the sqrt, it's valid to cancellate both symbols, the absolute value works only if the power is inside the sqrt Try getting the sqrt of different numbers and then square them, and you will see that there are no contradictions The problem of getting the sqrt of a square is that you first square, so the square function get to values to the same point but the sqrt only returns you to one of them, while getting the square of a sqrt is no problem because you do first the sqrt, which is a one to one function, and then you square it, and it gets you to only one point (because it's a function)
Where do you get "This square root function has the range zero to plus infinity" from? I was taught (when talking about positive real numbers) the square root function has 2 values, one positive and an equal negative one. You can't just take the positive value, you have to consider the negative one as well even if it eventually leads to a rejected result later. You even use this fact when you solve the binomial equation x = (-b PLUS or MINUS sqrt(b^2-4ac))/2a Hence x=0 is a valid solution (-1 = sqrt(1)) and if you substitute x=-0.618033989... (your other rejected root) x^2-1 = (-0.618...)^2 - 1 = -0.618... = sqrt(-0.618... +1) = sqrt(0.382...) = =/- 0.616... in other words x^2-1 = x = sqrt(x+1) when x = -0.618 if you take the negative square root.
First of all, there are two functions that mutually turn into one: x^2-1=(x+1)^1/2 => ((x-1)(x+1))^2=x+1 => (x-1)^2 (x+1)^2 - (x+1)=0 => (x+1)((x-1)^2 * (x+1) -1)= 0 => x =-1 and (x-1)^2 *(x+1) -1 =0 => x^3-2x^2+x+x^2-2x+1-1=0 => x^3 - x^2 - x = 0 => x(x^2-x-1)=0, so x = (1+5^1/2)/2. Answer: x=-1 and x=(1+5^1/2)/2
It's a polynomial division, you can only do it after you know a solution. The reasoning goes like that (and IMO should have been explained like that): If a polynomial with integer coefficients has an integer root, it must be a factor of the constant term, in this case -1. The only factors of -1 are +1 and -1. So we try both, and if we're lucky we have a solution. If not, we know that there is no integer solution, but otherwise we are not any closer to finding a solution. In this case, if we evaluate at +1, we obtain -2, thus +1 is not a solution. Then evaluate for -1 and we find that (-1)^3-2*(-1)-1=0, so -1 is indeed a solution. Thus it must be possible to divide the polynomial by x-(-1)=x+1. If we do this, we obtain a polynomial that has an order of 1 less so it becomes easier to calculate the remaining roots. The polynomial division can be done by adding zeros. The polynomial starts like x^3-2*x, but if I want to divide by x+1, it would be nice if it started with x^3+x^2, because that is x^2*(x+1). So therefore I add x^2, and then, in order to correct the error that I introduced, I need to subtract x^2 again. So then the remainder starts like -x^2-2*x, but I'd rather want -x^2-x, because that's -x*(x+1), so I split the -2*x into -x -x.
Your domain restriction was wrong. The only restriction on the domain is that the argument under the square root must be >=0 x+1>=0 x>=-1 x=0 is a perfectly good solution.
@@JayTemple 0^2-1 is negative, but that's not what's in the question, it's 0^2+1. And square roots are two values, a positive and an equal negative one.
(xx-1)(x-1)(x+1)=x+1 (xx-1)(x-1)=1 xxx-xx-x+1=1 xxx-xx-x=0 assuming x /= 0 ( not a solution) div by x xx=x+1 well known as the golden ratio x=(√5 −1)/2
x^2-1 =sqrt(x+1) so x^4 -2x^2 + 1 = x + 1 so x^4 - 2x^2 - x = 0 so x = 0 or x^3 - 2x - 1 = 0 so x^3 + x^2 - x^2 - x - x - 1 = 0 so (x+1)(x^2-x-1)=0 so x = -1 or x = (1+-sqrt5)/2 Checking we see that x = 0 is not a valid solution.
{ x²-1≥0 and x+1≥0 } or { |x|≥1 and x>-1 } or { (x≤-1 or x≥1 ) and x≥-1 } or x≥1. For x≥1 the given equation is equivalent to (x²-1)² =(√(x+1))² ... x³-2x-1=0 etc ...
The rejected root also makes L.S. = R.S. But, by the token X² - 1 > = 0 [ assuming √(x+1) to be a +ve value], the root like (1-√5)/2 is rejected. Really speaking, it is never a disqualifying root. Thus, there are 3 valid roots wherein x= 0 makes -1 = √1??!!
I don’t understand why x^2 - 1 >= 0. When you take square root for example 4 you get -2 and 2. Square roots can always be negative. 0 is an answer you just have to consider 1^(1/2) to be -1 which it is, since -1*-1 is always 1.
@petermaling943 The square root sign means positive root. However x^(1/2) could mean either. Why put +/- in the quadratic equation if the two roots are implied in the first place?
The square root sign almost always means the principal root, and this video is pretty clearly working in R so we shouldn't even be considering imaginary results, which negative square roots directly implicate.
@@decaydjk8922 Um 0 is a real root. I get the first thing you said but the end didn't make much sense. I'm not saying (-1)^(1/2) I'm saying 1^(1/2) can be -1.
But when we actually use X is equals to 0 we can still get the same answer Example x = 0 is true x² - 1 = (x+1)^1/2 0² - 1 = (0 + 1)½ -1 = 1½ -1 = √1 That is actually true Because √1 is can be -1 or 1 I think C.M.I.I.W
When using √ we only use principal (positive) square root. If you want negative value, put a negative sign in front of √. This is why quadratic formula *[x = (−b ± √(b²−4ac)) / (2a)]* has ± in front of √. Otherwise if √ could be positive or negative, we'd just use + in front of √. Examples: x = √4 → x = 2 x² = 4 → x = ±√4 = ±2
0 is a solution because 0^2-1=sqrt(0+1) -> -1= sqrt(1) ... -1 is a sqrt of 1 solution because sqrt implies both positive and negative numbers as long as x+1 >= 0 ... this is flawed math ideaology ... x>=-1 not x^2 >= 0 as the inequality.
@@slorattlesnake669 dear rage-typer, you are victim of a common misconception: The equation x²=1 has to real roots, since its second degree (fundamental theorem of algebra), which are x=1 and x=-1, but non because sqrt(1) is equal to ±1, but because the sqrt(x²), needing to be a positive numer, is equal to |x|. Therefore x²=1 can be written as |x|=sqrt(1), which translates in x=±sqrt(1)=±1. The equation x=sqrt(1) is linear and for the fundamental theorem of algebra it has only one root, which is only 1.
x = 0 gives a negative LHS, it must be positive. whereas it is a solution of the quartic it is NOT a solution of the given equation x = 1 gives 0 = √2 You have an error here RHS should be x +1
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
You should have mentioned that (1 + sqrt(5))/2 = φ, the Golden Ratio. It's a special number such that squaring it has the same result as adding 1 instead.
φ^2 = φ+1, and we can use this to swap the x^2 and (x+1) in the original equation:
(φ+1) - 1 = sqrt(φ^2)
φ = φ
if you set one side of the original eqn to y you would get a nice system of equations x=y^2-1 y=x^2-1. You can then use simple algebra and get (x-y)(x+y+1)=o This avoids having to solve a cubic like it was shown.
The lhs is (x+1)(x-1) so we immediately see x=-1 as a solution and can divide both sides by sqrt(x+1) before squaring. This gives the cubic equation.
Yes, exactly, if you recognize the left hand side as a difference of two squares you can simplify the equation without having to deal with the fourth degree polynomial.
Conclusion from this video: high school in the Netherlands has an atrocious quality. I never learned how to solve such a cubic equation. Adding some arbitrary X² and then finding a common factor to split it up. I see what he is doing, I am not sure about the logic behind it. It seems that the goal is to rewrite the equation in a form which is solvable by hand but how do you figure out how to get there? What if you wouldn't be able to get that common factor?
It's not jst abt this equation or abt this solution or abt the idea of this prob it is abt the logic abt the how you will get the idea? and i think the answer is the habit of doing math problems and there is not only this way to solve it what i want to say is u jst need to have the math logic and think of athor ways bcz most of tge times we don t get it from the frst try so we jst need to move the pen .
That's my question too. We see this manipulation all the time but it's never properly explained. Which is weird because they over-explain everything else but never this.
Cubic equations may be taken up by Trial & Error Method of factorisation in which +ve & -ve values like 1, 2, 3, ... are assigned to satisfy. Of course, for a root like x = -73.89 this method cannot be followed, rather by Sridhar Acharya or by computer to solve.
So the OP is a bit of an idiot. Look at x^3-2x-1. Plug in factors of the linear term - 1 and -1 - and we find (x+1) is a factor. So divide (x+1) and get (x^2-x-1).
The +x^2-x^2 trick is just a fancy way to divide by (x+1)
@@adamnevraumont4027 Do you feel better by insulting other people?
Es mejor pasar todo a un miembro de la ecuación y extraer factor común √(x+1). De hecho , creo que es la manera correcta, por su eficiencia, de resolver esta ecuación:
(x+1)(x-1)-√(x+1)=0
√(x+1) [(x-1)√(x+1) - 1]=0
Del primer factor obtenemos x=-1, que comprobamos y resulta ser válida.
Del segundo, elevamos al cuadrado en:
(x-1)√(x+1)=1
para obtener,
x(x²-x-1)=0
cuyas soluciones son:
x=0, que descartamos,
x=(1+√5)/2 (válida), y
x=(1-√5)/2, que también descartamos tras su comprobación.
Before watching the video:
One root is -1 (easy), the second is ~ +1.618 (calculated numerically)
When I divide the equation by the first root, I get (x-1)*SQRT(x+1)=1
Squaring that I get x³-x²-x=0 => x(x²-x-1)=0
x=0 does not work in the original equation
The quadratic equation has two roots 0.5*(1+/- SQRT(5))
The + variant works, the - variant does not.
=> The roots are -1 and 0.5*(1+SQRT(5)).
That's the way I did it too. It's also interesting that -1 no longer works as a solution to x^3-x^2-x=0 due to the earlier step of dividing by √(x+1), which constitutes division by zero when x=-1.
You can see from the graph that the two rejected "solutions" correspond to where x²-1 and √(x+1) are of opposite value. Which makes sense, as the square of a negative number is the same as the square of its positive counterpart, and our first manipulation of the equation was to square both sides...
Simplify x-1 = 1/(x+1)^0.5. Then square both sides. x^2 -2x + 1 = 1 / (x+1)
Transpose (x+1)(x^2 −2x+1)=1. Expand and simplify x^3−x^2−x=0 or x(x^2 −x−1)=0.
Roots. (1) x =0. Extraneous. By quadratic equation formula, (2) x = 1.618 acceptable ans. (3) x = - 0.618 acceptable ans.
(In case you don’t understand LaTeX notation, I’ve edited the comment and included ASCII equations next to the LaTeX expressions)
This problem is very elementary and *does not* require one to find the roots of a cubic polynomial.
The equation has two real roots of which one is obviously $-1$ ( -1 ). To find the positive real root, first, factor the LHS of the equation :
$(x + 1)(x - 1) = \sqrt{x + 1}$
( (x + 1)(x - 1) = sqrt(x + 1) )
Now square both sides :
$(x + 1)(x - 1)(x + 1)(x - 1) = (x + 1)$
( (x + 1)(x - 1)(x + 1)(x - 1) = (x + 1) )
Next, divide both sides by $(x + 1)$ ( (x + 1) ) :
$(x + 1)(x - 1)(x - 1) = 1 \ \text{for} \ x
e -1$
( (x + 1)(x - 1)(x - 1) = 1 , for x =/= -1 )
Now, expand the LHS :
$(x^{2} - 1)(x - 1) = x^{3} - x - x^{2} + 1 = 1$
( (x^2 - 1)(x - 1) = x^3 - x - x^2 + 1 = 1 )
Subtract $1$ ( 1 ) from both sides :
$x^{3} - x - x^{2} = 0$
( x^3 - x - x^2 = 0 )
Divide by $x$ ( x ) :
$x^{2} - 1 - x = 0$
( x^2 - 1 - x = 0 )
Solve for $x$ ( x ) :
$x = \dfrac{1}{2} + \dfrac{\sqrt{5}}{2}$
( x = 1/2 + sqrt(5)/2 )
let's solve from x³ - 2x - 1 = 0 and x² ≥ 1
x³ - x - (x + 1) = 0
x(x² - 1) - (x + 1) = 0
(x + 1)[ x(x -1) - 1 ]= 0
x + 1 = 0 ∨ x² - x - 1 = 0
x + 1 = 0 => *x = -1*
x² - x - 1 = 0
*x = (1 + √5)/2*
Yeah, 'saw that too.
Think of the implied geometry: recognize that what you are looking at is an equation asking for where a function and its inverse are equal. They are equal on the line y = x. The solution is then very simple as you can solve x^2 - 1 = x or sqrt( x + 1 ) = x. My guess is that this is the intended method of solution as it is not tedious.
My thoughts exactly
Your method yields 2 solutions, of which, one doesn't work. And it misses another true solution.
@@oahuhawaii2141 if you think of the graphs, the -1 solution is obvious and the positive solution follows this rule. You can even easily determine which of the two solutions is rejected. (I was able to solve in my head because of this.)
@@gregorylewis4426: But the graph isn't a solution. Rather, it merely shows where the solutions lay, and they still must be found. The solution OP gave is lacking, and simply saying "it's easily determined" isn't going to cut it on an exam: I didn't solve the problem because it's obvious or easily found.
@@oahuhawaii2141More formally, for negative x values, you have a monotonically decreasing function and a monotonically increasing function, so there can only be one intersection, and we can easily see that the x value for that is -1. For the non-negative half, we have a function and its inverse, so we can set x=y. From there we have x^2-1=x, so x^2-x-1=0, so x=(1+-√5)/2, and since (1-√5)/2 is negative, we reject that, giving us our two solutions. Neither solution is better. This one is just much simpler.
A and B beeing two reals, we have: sqrt(A) = B is equivalent to A = B^2 and B>=0.
So the given equation is equivalent to (x^2 -1)^2 = x + 1 and x^2 - 1 >=0 (i.e. x is in E = ]-infinity, -1] U [1, +infinity[ )
We solve (x^2 -1)^2 = x + 1, and among the values of x that we obtain, the solutions of the given equation are the ones beeing in E.
There is a slightly easier way to solve this, where we don't have the problem to solve a cubic equation.
Starting with squaring both sides we get
(x^2-1)^2=x+1 Now we use the second binomial formula on the left side to factor :
(x-1)^2*(x+1)^2=x+1. We see both sides have the factor x+1. We check if x+1=0 is a solution, which is indeed the case. Therefor x1=-1.
Case 2: x-1, that means x+10. In this case we can divide both sides of the equation by x+1.
We get (x+1)(x-1)^2=1. Now we expand
(x+1)(x^2-2x+1)=x^3-2x^2+x+x^2-2x+1=x^3-x^2-x+1=1 Subtracting 1 on both sides gives us
x^3-x^2-x=0. We can now easily see that we can factor x:
x(x^2-x-1)=0. We check x=0 in the original equation and see it is not a solution. Therefor we can divide both sides by x:
x^2-x-1=0 It's the same quadratic equation like in the video which gives us the same solution, the (1-sqrt(5))/2 we reject, because checking in the original equation reveals that this is not a solution.
2 things to note: if you square both sides of an equation you always have to check if your solutions are true solutions because this step can give you extraneous solution, which indeed happened in this case.
And second if you divide both sides of an equation by a term containing x, always check, if this term can equal zero. Because dividing both sides of an equation by zero is not allowed and in this case you can lose solutions. Would we have just divided by x+1 without checking, we would have lost the solution -1.
Square both sides
(x+1)(x-1)(x-1)(x+1)=(x+1)
Set x+1=0, x=-1
Divide both sides by x+1
(X2-1)(X-1)=1=X3-x2-x+1
X3-x2-x=0=x(x2-x-1)
X=-1,0,(1+-{5})/2 but 0 and (1-{5})/2 are artifacts from squaring both sides
You can only divide by (x+1) if it’s =/= 0 so in this case x=/=-1
@@wanou_4259: Uh, it's the same as factoring, and setting each factor to 0.
@@oahuhawaii2141 true, but dividing by x+1 then saying -1 is an answer is wrong since it would result in dividing by 0
@@wanou_4259: Well, the long and winding way to do it is:
LHS: x² - 1 ≥ 0 ; |x| ≥ 1 ; x ≥ 1 or -1 ≥ x
RHS: √(x + 1) ≥ 0 ; x ≥ -1
Therefore, x ≥ 1 or x = -1 .
x² - 1 = √(x + 1)
(x + 1)*(x - 1) - √(x + 1) = 0
√(x + 1)*(√(x + 1)*(x - 1) - 1) = 0
Left factor:
√(x + 1) = 0
x = -1 -- first solution
Right factor:
√(x + 1)*(x - 1) - 1 = 0
√(x + 1)*(x - 1) = 1
(x + 1)*(x - 1)² = 1
(x² - 1)*(x - 1) = 1
x³ - x² - x + 1 = 1
x*(x² - x - 1) = 0
x = 0, (1 ± √5)/2 -- accept only x ≥ 1
x = (√5 + 1)/2 -- second solution
That was really teaching thank you sir
Which software is used while thumbnail making please. And how you write math eqution.
Good work, but since x = -1 is a trivial solution easily found by plug and check, we could reduce the amount of work by dividing the 4th degree equation down to a cubic right off the bat.
√x² = x is not an identity. Use this substitution in an equation is a heavy error. Use this substitution in an educational video is uneducational..
√x² = |x| is an identity, good for all values of x.
Well yes sqrt x^2= abs(x) this is most likely for high schoolers who haven't learned what an absolute value is so 🤷
@ So you explain high schooler a *wrong* stuff? A wrong thing they remember forever, causing them to make mistakes in the future?
I don’t think so.
I think it’s a limit of the author’s teaching abilities.
Obviously in this case the step works, the two equations (before and after the substitution) are equivalents, but only for the specific properties of this case. You have to specify this, if you love your students.
When does he says that? If the power is outside the sqrt, it's valid to cancellate both symbols, the absolute value works only if the power is inside the sqrt
Try getting the sqrt of different numbers and then square them, and you will see that there are no contradictions
The problem of getting the sqrt of a square is that you first square, so the square function get to values to the same point but the sqrt only returns you to one of them, while getting the square of a sqrt is no problem because you do first the sqrt, which is a one to one function, and then you square it, and it gets you to only one point (because it's a function)
@
1. “when does he say it?”
Minute 1.20
2. The rest
I know, that is what I said.
@ZannaZabriskie you mean he should have said that it only works becuase the square is outside the sqrt?
x² - 1 = √(x + 1)
LHS: |x| ≥ 1 ; RHS: x ≥ -1 ; x ≥ 1 or x = -1
(x - 1)*(x + 1) = √(x + 1)
Factor out √(x + 1) and set to 0 -> x = -1 .
Continue working on the equation:
(x - 1)*√(x + 1) = 1
(x - 1)²*(x + 1) = 1
(x² - 1)*(x - 1) = 1
x³ - x² - x + 1 = 1
x*(x² - x - 1) = 0
x = 0, (1 ± √5)/2
∴ x = -1, (√5 + 1)/2
Where do you get "This square root function has the range zero to plus infinity" from?
I was taught (when talking about positive real numbers) the square root function has 2 values, one positive and an equal negative one. You can't just take the positive value, you have to consider the negative one as well even if it eventually leads to a rejected result later.
You even use this fact when you solve the binomial equation x = (-b PLUS or MINUS sqrt(b^2-4ac))/2a
Hence x=0 is a valid solution (-1 = sqrt(1))
and if you substitute x=-0.618033989... (your other rejected root) x^2-1 = (-0.618...)^2 - 1 = -0.618... = sqrt(-0.618... +1) = sqrt(0.382...) = =/- 0.616...
in other words x^2-1 = x = sqrt(x+1) when x = -0.618 if you take the negative square root.
❶ x = ½ (1 -√5) ✘ as x² -1 ≠ √(1 +x),
❷ x = ½ (1 +√5) ✔ as x² -1 = √(1 +x). 👈
From LHS x²-1>=0 --> x²>=1
|x|>=1
From RHS x>=-1
Therefore x>=-1
x3 is golden ratio. Clearly x3>1
Clearly x42
If this appears on the SAT you can just use desmos
An equation of 4 th degree has at most 4 solutions, not " at least " .( Fundamental theorem of algebra).
x^2-1=rt(x+1)
x>1 or x
First of all, there are two functions that mutually turn into one: x^2-1=(x+1)^1/2 => ((x-1)(x+1))^2=x+1 => (x-1)^2 (x+1)^2 - (x+1)=0 => (x+1)((x-1)^2 * (x+1) -1)= 0 => x =-1 and (x-1)^2 *(x+1) -1 =0 => x^3-2x^2+x+x^2-2x+1-1=0 => x^3 - x^2 - x = 0 => x(x^2-x-1)=0, so x = (1+5^1/2)/2. Answer: x=-1 and x=(1+5^1/2)/2
√(x+1)=0, x1=-1✓
(x+1-2)√(x+1)=1
a³-2a-1=0
a³+1-2(a+1)=0
(a+1)(a²-a-1)=0
a1=-1, a⅔=(1±√5)/2
√(x+1)=(1+√5)/2
x+1=(3+√5)/2
x=(3+√5-2)/2=(1+√5)/2✓
x=(-1, (1+√5)/2)
x²-1=sqrt(x+1)
(x+1)(x-1)=sqrt(x+1)
[sqrt(x+1)]²(x-1)=sqrt(x+1)
[sqrt(x+1)][{(x-1)sqrt(x+1)}-1]=0
sqrt(x+1)=0 --> x=-1
[sqrt(x+1)](x-1)=1
sqrt(x+1)=1 --> x=0 rejected
x-1=1 --> x=2 rejected
x=-1 is solution of x^3-2x-1=0 so
x^3-2x-1=(x+1)(x^2-x-1)
I just don’t understand why zero is rejected, isn’t-1 a square root of 1 since -1 x -1 = 1?
Can you generalize you 0-additon trick?
No, does not work in general.
It's a polynomial division, you can only do it after you know a solution. The reasoning goes like that (and IMO should have been explained like that):
If a polynomial with integer coefficients has an integer root, it must be a factor of the constant term, in this case -1. The only factors of -1 are +1 and -1. So we try both, and if we're lucky we have a solution. If not, we know that there is no integer solution, but otherwise we are not any closer to finding a solution. In this case, if we evaluate at +1, we obtain -2, thus +1 is not a solution. Then evaluate for -1 and we find that (-1)^3-2*(-1)-1=0, so -1 is indeed a solution. Thus it must be possible to divide the polynomial by x-(-1)=x+1. If we do this, we obtain a polynomial that has an order of 1 less so it becomes easier to calculate the remaining roots.
The polynomial division can be done by adding zeros. The polynomial starts like x^3-2*x, but if I want to divide by x+1, it would be nice if it started with x^3+x^2, because that is x^2*(x+1). So therefore I add x^2, and then, in order to correct the error that I introduced, I need to subtract x^2 again. So then the remainder starts like -x^2-2*x, but I'd rather want -x^2-x, because that's -x*(x+1), so I split the -2*x into -x -x.
x²-1=√(x+1)
Domain: x≤-1, x≥1
x⁴-2x²+1=x+1
x⁴-2x²=x
x⁴-2x²-x=0
x(x³-2x-1)=0
x=0 ∅ see domain
x³-2x-1=0
x³-x-x-1=0
x(x²-1)-1(x+1)=0
x(x+1)(x-1)-1(x+1)=0
(x+1)[x(x-1)-1]=0
(x+1)(x²-x-1)=0
x²-x-1=0
4x²-4x-4=0
4x²-4x+1=5
(2x-1)²=5
|2x-1|=√5
2x-1=-√5
2x=1-√5
2x≈-1.24
x≈-0.62 ∅ see domain
2x-1=√5
2x=1+√5
x=½(1+√5) ❤
x+1=0
x=-1 ❤
Your domain restriction was wrong.
The only restriction on the domain is that the argument under the square root must be >=0
x+1>=0
x>=-1
x=0 is a perfectly good solution.
But 0^2 - 1 is negative, and square roots are never negative.
@@JayTemple Oh damn you're right
@@andrewstallard6927
No you are right, it's x^2+1 under the square root, not x^2-1
@@JayTemple 0^2-1 is negative, but that's not what's in the question, it's 0^2+1.
And square roots are two values, a positive and an equal negative one.
@@pjaj43 Yes, but the square root sign itself refers exclusively to the principal (i.e., positive) one.
I lost you at D=b squared - 4ac
Where did this come from?
Bhaskara
x^2-1=(x+1)(x-1)= rt(x+1)
Square both side
(x+1)^2*(x-1)^2= x+1
(x+1)((x+1)(x-1)^2-1)
(x+1)((x^2-1)(x-1)-1)
(x+1)(x^3-x^2-x)
...
x>=1 ,x=-1
Then x>=1, x=-1
(xx-1)(x-1)(x+1)=x+1
(xx-1)(x-1)=1
xxx-xx-x+1=1
xxx-xx-x=0
assuming x /= 0 ( not a solution) div by x
xx=x+1 well known as the golden ratio
x=(√5 −1)/2
x+1 > x
x^2-1 =sqrt(x+1)
so x^4 -2x^2 + 1 = x + 1
so x^4 - 2x^2 - x = 0
so x = 0 or
x^3 - 2x - 1 = 0
so x^3 + x^2 - x^2 - x - x - 1 = 0
so (x+1)(x^2-x-1)=0
so x = -1
or x = (1+-sqrt5)/2
Checking we see that x = 0 is not a valid solution.
X=0
x^2-1=√x+1 =(x+1)(x-1)=(x+1)^1/2
(x+1)(x-1)÷(x+1)=(x+1)^1/2 ÷(x+1)
(x-1)=(x+1)^1/2 . (x+1)^-1
=(x+1)^-1/2
(x+1)^2 = (x+1)
x^2 +2x +1 =x +1
x^2 + 2x - x +1 - 1=0
x^2. + x=0
x(x+1). =0
either x=0 or (x+1)=0
therefore x+1=0. ; x=-1
Test if x=0. x^2 -1~√x+1. Test: if x=1
(- 1)^2 -1~√-1 +1
1 -1~ 0
therefore 0=0 ; if x= -1
0 -1~√0+1
-1~1
reject
{ x²-1≥0 and x+1≥0 } or
{ |x|≥1 and x>-1 } or
{ (x≤-1 or x≥1 ) and x≥-1 } or x≥1.
For x≥1 the given equation is equivalent to (x²-1)² =(√(x+1))²
... x³-2x-1=0 etc ...
nowosc czas na
The rejected root also makes L.S. = R.S.
But, by the token X² - 1 > = 0
[ assuming √(x+1) to be a +ve value], the root like (1-√5)/2 is rejected. Really speaking, it is never a disqualifying root. Thus, there are 3 valid roots wherein x= 0 makes -1 = √1??!!
let y = sqrt (x+1)
then y > = 0 , y > = - 1
finally y > = 0
now y^2 = x+1 , y = x^2 -1
y^2+y = x^x+x
(y-x) (y+x+1) = 0
(y - y^2+1) (y+y^2) = 0
y (y+1)(y^2-y-1) = 0
y = 0 , -1 , (1+√5)/2 , (1 - √5)/2
now y > = 0 hence
y = 0 , (1+√5)/2
and x = y^2 - 1 = - 1 , (1+√5)/2
х=(-1)
Is it x is zero
х=0 х=-1
Is the tricky part still with us?
-1
x(x^3+/-x^2-2x-1)=0 , /x=0 , not a solu /, x^3+/-x^2-2x-1=0 , (x-1)(x^2-x-1)=0 ,
1 1
-1 -1
-1 -1
I don’t understand why x^2 - 1 >= 0. When you take square root for example 4 you get -2 and 2. Square roots can always be negative. 0 is an answer you just have to consider 1^(1/2) to be -1 which it is, since -1*-1 is always 1.
The square root sign implies just the positive root. Otherwise we would add the +/- sign.
@@mcwulf25 No, it doesnt. The commenter is correct and the video is wrong.
@petermaling943 The square root sign means positive root. However x^(1/2) could mean either. Why put +/- in the quadratic equation if the two roots are implied in the first place?
The square root sign almost always means the principal root, and this video is pretty clearly working in R so we shouldn't even be considering imaginary results, which negative square roots directly implicate.
@@decaydjk8922 Um 0 is a real root. I get the first thing you said but the end didn't make much sense. I'm not saying (-1)^(1/2) I'm saying 1^(1/2) can be -1.
(x^2)^2 ➖ (1)^2 {x+x ➖ }+{1+1 ➖ }{x^4 ➖ 1} {x^2+2}={x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖ x^0+x^0 ➖} 2x^2={x^1+x^1+x^1+x^1}(x ➖ 2x+2) =x^4 x^2^2 (x ➖ 2x+2).
ESPAÑA
But when we actually use X is equals to 0 we can still get the same answer
Example x = 0 is true
x² - 1 = (x+1)^1/2
0² - 1 = (0 + 1)½
-1 = 1½
-1 = √1
That is actually true
Because √1 is can be -1 or 1
I think
C.M.I.I.W
Definition of √1 is only the positive number.
When using √ we only use principal (positive) square root. If you want negative value, put a negative sign in front of √. This is why quadratic formula *[x = (−b ± √(b²−4ac)) / (2a)]* has ± in front of √. Otherwise if √ could be positive or negative, we'd just use + in front of √.
Examples:
x = √4 → x = 2
x² = 4 → x = ±√4 = ±2
0 is a solution because 0^2-1=sqrt(0+1) -> -1= sqrt(1) ... -1 is a sqrt of 1 solution because sqrt implies both positive and negative numbers as long as x+1 >= 0 ... this is flawed math ideaology ... x>=-1 not x^2 >= 0 as the inequality.
Boy u must be dumb
No 0 is not a solution... square root can never be equal to a negative number
@@Bandicoot_Power (-1)x(-1) = 1 are you delusional?
sqrt(1)= 1 or -1
@@slorattlesnake669 dear rage-typer, you are victim of a common misconception:
The equation x²=1 has to real roots, since its second degree (fundamental theorem of algebra), which are x=1 and x=-1, but non because sqrt(1) is equal to ±1, but because the sqrt(x²), needing to be a positive numer, is equal to |x|. Therefore x²=1 can be written as |x|=sqrt(1), which translates in x=±sqrt(1)=±1.
The equation x=sqrt(1) is linear and for the fundamental theorem of algebra it has only one root, which is only 1.
Square both sides of the equation: x^4 - 2x^2 + 1 = x -1: Simplify and factor: use 0 factor theorem to get x = 0, +1, -1. All check out.
x = 0 gives a negative LHS, it must be positive. whereas it is a solution of the quartic it is NOT a solution of the given equation
x = 1 gives 0 = √2 You have an error here RHS should be x +1
X=0 is not a solution because -1=1 is wrong !
Did you watch the video? LOL
@janwip Please excuse. I habve red too fast. I am very sorry.
No -1 = -1 so 0 is a solution. Note sqrt(1) = +/-1 and -1 satisfies the equation
x² - 1 = √(x + 1)
[ x² ≥ 1 ]
x + 1 = u²
x² - 1 = √u² = ± u
case 1
x + 1 = u²
u + 1 = x²
x² - u² = -x + u
(x + u)(x - u) = - (x - u)
(x - u)(x + u + 1) = 0
u = x ∨ u = - (x + 1)
case 1.1
u = x
x² - x - 1 = 0
*x = (1 + √5)/2*
case 1.2
u = - (x + 1)
x² + x = 0 => x(x + 1) = 0
*x = -1*
case 2
x + 1 = u²
-u + 1 = x²
x² - u² = - (x + u)
(x + u)(x - u) + (x + u) = 0
(x + u)(x - u + 1) = 0
u = -x ∨ u = x + 1
case 2.1
u = -x
x² - x - 1 = 0 [ like case 1.1 ]
case 2.2
u = x + 1
x² + x = 0 [ like case 1.2 ]
√u² = ± u FALSE
√u² = |u| TRUE
I find it so distracting that the x is written using a backward c followed by c. Couldn't follow the lesson.
on s'en tape de toute façon le dialogue n'est que dans un sens et tu fais toi meme beaucoup de fautes
Quelles fautes?