Nice solution. Other approach to the equation x^2-3=sqrt(x+3): After substitution a=3 we get x^2-a=sqrt(x+a). After squaring both sides we get x^4-2ax^2+a^2=x+a, i.e. a^2-(x^2+1)a+x^4-x=0 which is quadratic in a with delta=(2x^2+1)^2-x^4+4x=(2x+1)^2 and roots a1=(2x^2+1+2x+1)/2=x^2+x+1 and a2=(2x^2+1-2x-1)/2=x^2-x. After back substitution we get two quadratics in x: x^2+x+1=3 and x^2-x=3 i.e. x^2+x-2=0 and x^2-x-3=0 as presented in video.
I came up with my own method similar to your first method but for me it was clearer: After squaring the original eqn and combining terms I got x^4 - 6x^2 - x + 6 = 0. I assumed this can be written in the form (x^2 - a)^2 - (x - b)^2 (difference of squares). I multiplied these out and solved for a and b, which gives a = 5/2, b = -1/2. The equation then factors into [(x^2 - 5/2) - (x+1/2)] [(x^2 - 5/2) + (x + 1/2)] = (x^2 - x - 3) (x^2 + x - 2). I factored these terms by the quadratic eqn and simple factorization to get the solutions. Nice problem.
Bravo, this is an equation that is difficult to find even one way to solve. Thanks to your method, I found the solution to the following equation: x^2-13=(x+13)^1/2
I have found a method which although incomplete I find very nice. By moving 3 to the right hand side: x^2=3+sqrt(3+x) So: x=sqrt(3+sqrt(3+x)) By looking at this we notice the RHS looks like a composition of a function f with itself. Define: f(x)=sqrt(3+x) RHS=f(f(x)) LHS=x Therefore: f(f(x))=x f(x)=f^-1(x) And we know a function and its inverse will meet at y=x: So: f^-1(x)=x By inspection of the original eq: f^-1(x)=x^2-3 So: x^2-3=x We get one sol with the quadratic formula. I don't know where the second sol was lost.
Thankyou for sharing.
My pleasure
Thanks 🙏🙏🙏🙏
You’re welcome 😊
Thank you its usefull.
Glad it was helpful!
Nice solution.
Other approach to the equation x^2-3=sqrt(x+3):
After substitution a=3 we get x^2-a=sqrt(x+a).
After squaring both sides we get x^4-2ax^2+a^2=x+a, i.e. a^2-(x^2+1)a+x^4-x=0 which is quadratic in a with delta=(2x^2+1)^2-x^4+4x=(2x+1)^2 and roots a1=(2x^2+1+2x+1)/2=x^2+x+1 and a2=(2x^2+1-2x-1)/2=x^2-x.
After back substitution we get two quadratics in x: x^2+x+1=3 and x^2-x=3 i.e. x^2+x-2=0 and x^2-x-3=0 as presented in video.
I came up with my own method similar to your first method but for me it was clearer:
After squaring the original eqn and combining terms I got x^4 - 6x^2 - x + 6 = 0.
I assumed this can be written in the form (x^2 - a)^2 - (x - b)^2 (difference of squares).
I multiplied these out and solved for a and b, which gives a = 5/2, b = -1/2.
The equation then factors into [(x^2 - 5/2) - (x+1/2)] [(x^2 - 5/2) + (x + 1/2)] = (x^2 - x - 3) (x^2 + x - 2).
I factored these terms by the quadratic eqn and simple factorization to get the solutions.
Nice problem.
Nice👍
Thanks!
Bravo, this is an equation that is difficult to find even one way to solve.
Thanks to your method, I found the solution to the following equation: x^2-13=(x+13)^1/2
Glad to hear that!
I have found a method which although incomplete I find very nice.
By moving 3 to the right hand side:
x^2=3+sqrt(3+x)
So:
x=sqrt(3+sqrt(3+x))
By looking at this we notice the RHS looks like a composition of a function f with itself.
Define:
f(x)=sqrt(3+x)
RHS=f(f(x))
LHS=x
Therefore:
f(f(x))=x
f(x)=f^-1(x)
And we know a function and its inverse will meet at y=x:
So:
f^-1(x)=x
By inspection of the original eq: f^-1(x)=x^2-3
So:
x^2-3=x
We get one sol with the quadratic formula.
I don't know where the second sol was lost.
The x = 1 solution of the quartic can be seen much easier than the x = -2 solution ( without rational root theorem).
Synthetic division saves you the problem of juggling the coefficients.
Yes but juggling the coefficients is more fun! 🤪
x^2 - 3 = sqrt(x+3)
x^4 - 6*x^2 + 9 = x + 3
x^4 - 6*x^2 - x + 6 = 0
(x-1)*(x+2)*(x^2 - x - 3) = 0
x = 1, x = -2, x=0.5+sqrt(13)/2, x=0.5-sqrt(13)/2.
Pardon but x-1 will be the 1st factor not x-2
x - 1 and x + 2 are factors.