I substituted y = sqrt(x); this factored to (y+1)(y-2)(y^2 + 2) which was a little easier to factor than in the first method; same answers. Note that x = -2 does solve the original equation if complex numbers are allowed; but gives a different value for x^2 - 5x
I multiplied the whole equation by x (not equal to 0), which results in: x*sqrt(x)+4 = x^2 - 2sqrt(x). Then bring all the sqrt(x)-terms to the left side and all other terms to the right side: x*sqrt(x) + 2sqrt(x) = x^2 - 4 Now factor out sqrt(x) on the left side and factor the right side according to difference of two squares: sqrt(x)*(x + 2) = (x - 2)*(x + 2) Since x cannot equal -2, we can divide by (x+2): sqrt(x) = x - 2 Squaring this equation gives us: x = x^2 - 4x + 4 Solving for x^2 gives: x^2 = 5x - 4 Now plug in the term of x^2 into the goal term: x^2 - 5x = (5x - 4) -5x = - 4. This comes close to your 2nd method, but with slightly different details. And without explicitly solving for x.
@ Ralf Bodemann -- No, back up your work five lines to this (after my edits): Squaring both sides of the equation gives: x = x^2 - 4x + 4. Solving for x^2 - 5x in terms of a constant gives: x^2 - 5x = -4. You're done at that point.
you missed the obvious one at 2:55 with x(sqrtx)+2(sqrtx) = x^2-4. Instead of squaring both sides, as you did, factor out the (x+2) from both sides to get (x+2)(sqrtx)=(x+2)(x-2). Then (assuming x not -2) square both sides to get x=x^2-4x+4 and subtract x for 0=x^2-5x+4 which leads very nicely to x^2-5x = -4
I also used the second method but instead of substituting in the equation X-2=sqrt(X), I squared both sides to get X^2-4X +4=X, which (when reorganized) gives x^2-5X=-4.
Like you, I gathered all the square-roots of x on one side, but I didn't see the factoring of the side w/o the square-root into product of sum&diff. Darn. Instead, I squared both sides, getting x + 4/x + 4 = x^2 + 16/x^2 -8. Then, I let u = x + 4/x, so the squared equation becomes: u + 4 = (u^2 - 8) - 8 = u^2 - 16. The rest is finding u, then x. tossing the extraneous candidate sol'ns, then evaluating the final expression. A LOT longer than your Method 2, but at least I got the same answer (-4). I'm learning, from all your daily math problems, that THE single, most important, general technique in Algebra is... a judicial SUBSTITUTION!! It keeps coming up! I'm learning to use it a LOT more, thx to you.
Very nice! I solved it using the second method ! I saw it almost immediately!! Not the very obvious one but the mind is mysterious some times!!😂 thank you!!!!❤
Isn't method 3 just an extra step added to method 1? Since you have x^2 - 4, you can factor out an sqrt(x) on the left side and expand the right to (x-2)(x+2). x+2 cancels out and you have sqrt (x) = x - 2.
It s also possible put t =sqrt(x), and then resolve t^2 - 2/t - 4/t^2 - t = 0. Thus (2/t +t) (t - 2/t - 1) = 0 => only (t - 2/t - 1) = 0 cause t>0. Just one positive good solution is t = 2..
sqrt x + 2/sqrt x = x -4/x ( add 2/sqrt x on both sides then subtract -4/x from both sides) x+ 4/x + 4 = x^2 + 16/x^2 -8 (square both sides) x+4/x + 20 = x^2 + 16/x^2 +8 ( add 16 to both sides of the equation) x+4/x + 20 = (x+4/x)^2 (factor the right side) lent n= x+4/x then n + 20 = n^2 n^2-20-n=0 (n-5)(n+4)=0 n=5 and n=-4: hence x+4/x = 5 and x+4/x = -4 For x+4/x =5, x^2 +4 = 5x [multiplied both sides by x] x^2-5x+4 =0 (x-4)(x-1)=0 x=4 and x=1 For x+4/x =-4 x^2+ 4 = - 4x [ multiplied both sides by x] x^2 +4x+4 =0 (x+2)(x+2)=0 x=-2 So for the equation x = 4, 1, and - 2 let's put all values of x into the original equation: sqrt x + 4/x =x- 2/sqrt x ruled out -2 since it will get the square root of a negative number For "1" you get 1 + 4/1 = 1- 2/1 5 =-1 didn't work For "4" you get 2 + 1 = 4- 1 3 =3 : hence x =4 hence x^2 - 5x = 16-20=-4 Answer
2:58 When you had xsqrtx+2sqrtx=x^2-4. sqrtx(x+2)=(x+2)(x-2) x>0 so ➗️ x+2 both sides sqrtx=x-2 then sq both sides you get x=1 and x=4. x=1 doesn't work x=4 is the only solution.
√x + 2/√x = x - 4/x Because the left-hand side is always positive, the right-hand side should be too… Squaring both sides yields an extra condition for real solutions, namely that x - 4/x >= 0. If we solve this inequality, we get x² >= 4 which results in -2>=x and x>=2. Because of the first condition (x>=0) the final quadrature condition becomes x >= 2 (x-1)(x-4)(x+2)²=0 This way there’s only one real root and that’s 4.
Why is the second part of the problem even there if we're going to solve for x in every method? How about expanding to the quartic, and then creating two quadratics: x^2-5x-a, and x^2+bx+c. Multiply them together and set the coefficients of each power of x to the ones in the quartic. b and c are easy to solve for (you don't even need c) and then a is your answer. Skip solving for x.
I used the second method. But I didn't understand why you want to evaluate x^2-5x instead of simply solving the equation for x. Probably to make it more tricky?😁
Try to find common factor. sqrt(x)+2'sqrt(x)=x-4/x, sqrt(x)+2/sqrt(x)=(sqrt(x)+2/sqrt(x))(sqrt(x)-2/sqrt(x))). So sqrt(x)+2/sqrt(x)=0 or 1=sqrt(x)-2/sqrt(x). x=4
sqrt(x) + 4/x = x - 2/sqrt(x) x•sqrt(x) + 4 = x^2 - 2•sqrt(x) x•sqrt(x) + 2•sqrt(x) = sqrt(x)•(x + 2) = x^2 - 4 = (x - 2)•(x + 2) x + 2 = 0 or sqrt(x) = x - 2 x = -2 or x - sqrt(x) - 2 = sqrt(x)^2 - sqrt(x) - 2 = (sqrt(x) - 2)•(sqrt(x) + 1) = 0 FALSE or sqrt(x) - 2 = 0 x = 4. x = 4 ==> x^2 - 5•x = 4^2 - 5•4 = 16 - 20 = -4. The comments have been saying that x = -2 is possible if complex numbers are allowed. However, this is inaccurate. The radical symbol is not well-defined for complex numbers. It is only well-defined for ordered fields which are square-root closed.
Nice! 14 can also be the answer if we consider complex numbers than x=-2 is a solution sqrt(-2)+4/-2=sqrt2 i -2=-2-2/(sqrt2 i)=sqrt2 i -2.
Interesting
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
x = -2 also works
I substituted y = sqrt(x); this factored to (y+1)(y-2)(y^2 + 2) which was a little easier to factor than in the first method; same answers. Note that x = -2 does solve the original equation if complex numbers are allowed; but gives a different value for x^2 - 5x
Pretty good
i did the exact same. i'm confused why x=-2 is not a valid answer, which produces a second answer to this problem, 14.
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
I multiplied the whole equation by x (not equal to 0), which results in: x*sqrt(x)+4 = x^2 - 2sqrt(x).
Then bring all the sqrt(x)-terms to the left side and all other terms to the right side: x*sqrt(x) + 2sqrt(x) = x^2 - 4
Now factor out sqrt(x) on the left side and factor the right side according to difference of two squares: sqrt(x)*(x + 2) = (x - 2)*(x + 2)
Since x cannot equal -2, we can divide by (x+2): sqrt(x) = x - 2
Squaring this equation gives us: x = x^2 - 4x + 4
Solving for x^2 gives: x^2 = 5x - 4
Now plug in the term of x^2 into the goal term: x^2 - 5x = (5x - 4) -5x = - 4.
This comes close to your 2nd method, but with slightly different details.
And without explicitly solving for x.
This is what I was thinking initially but then I lost it 😂
That's what I did. Way easier. Too bad you missed it SM. We all miss those things from time to time so don't feel bad. Really like your channel🖖
@ Ralf Bodemann -- No, back up your work five lines to this (after my edits):
Squaring both sides of the equation gives: x = x^2 - 4x + 4.
Solving for x^2 - 5x in terms of a constant gives: x^2 - 5x = -4.
You're done at that point.
I used the second method with a slightly more circuitous route.
you missed the obvious one at 2:55 with x(sqrtx)+2(sqrtx) = x^2-4. Instead of squaring both sides, as you did, factor out the (x+2) from both sides to get (x+2)(sqrtx)=(x+2)(x-2). Then (assuming x not -2) square both sides to get x=x^2-4x+4 and subtract x for 0=x^2-5x+4 which leads very nicely to x^2-5x = -4
I also used the second method but instead of substituting in the equation X-2=sqrt(X), I squared both sides to get X^2-4X +4=X, which (when reorganized) gives x^2-5X=-4.
I also did like this only
Ammazing th-cam.com/video/hozq83kmgbc/w-d-xo.html
Like you, I gathered all the square-roots of x on one side, but I didn't see the factoring of the side w/o the square-root into product of sum&diff. Darn. Instead, I squared both sides, getting x + 4/x + 4 = x^2 + 16/x^2 -8. Then, I let u = x + 4/x, so the squared equation becomes: u + 4 = (u^2 - 8) - 8 = u^2 - 16. The rest is finding u, then x. tossing the extraneous candidate sol'ns, then evaluating the final expression. A LOT longer than your Method 2, but at least I got the same answer (-4).
I'm learning, from all your daily math problems, that THE single, most important, general technique in Algebra is... a judicial SUBSTITUTION!! It keeps coming up! I'm learning to use it a LOT more, thx to you.
Very nice! I solved it using the second method ! I saw it almost immediately!! Not the very obvious one but the mind is mysterious some times!!😂 thank you!!!!❤
Excellent! Np and thank you
x = 4
16-20=-4
In second method when you had x-2=sqrtx just square both sides then x^2-4x +4=x
x*2-5x=-4 saves a little bit of time.
I solved in mind by third method but without t. Just squared both parts of eq and get the answer strightforward without unwanted root.
Isn't method 3 just an extra step added to method 1? Since you have x^2 - 4, you can factor out an sqrt(x) on the left side and expand the right to (x-2)(x+2). x+2 cancels out and you have sqrt (x) = x - 2.
It s also possible put t =sqrt(x), and then resolve t^2 - 2/t - 4/t^2 - t = 0. Thus (2/t +t) (t - 2/t - 1) = 0 => only (t - 2/t - 1) = 0 cause t>0. Just one positive good solution is t = 2..
sqrt x + 2/sqrt x = x -4/x ( add 2/sqrt x on both sides then subtract -4/x from both sides)
x+ 4/x + 4 = x^2 + 16/x^2 -8 (square both sides)
x+4/x + 20 = x^2 + 16/x^2 +8 ( add 16 to both sides of the equation)
x+4/x + 20 = (x+4/x)^2 (factor the right side)
lent n= x+4/x then
n + 20 = n^2
n^2-20-n=0
(n-5)(n+4)=0
n=5 and n=-4: hence
x+4/x = 5 and x+4/x = -4
For x+4/x =5,
x^2 +4 = 5x [multiplied both sides by x]
x^2-5x+4 =0
(x-4)(x-1)=0
x=4 and x=1
For x+4/x =-4
x^2+ 4 = - 4x [ multiplied both sides by x]
x^2 +4x+4 =0
(x+2)(x+2)=0
x=-2
So for the equation x = 4, 1, and - 2
let's put all values of x into the original equation: sqrt x + 4/x =x- 2/sqrt x
ruled out -2 since it will get the square root of a negative number
For "1" you get 1 + 4/1 = 1- 2/1
5 =-1 didn't work
For "4" you get 2 + 1 = 4- 1
3 =3 : hence x =4
hence x^2 - 5x = 16-20=-4 Answer
Nice
I miss the graph.
Sorry.
👍 / second method is a smartest one
I think so too!
I used Method 0. Look at the equation, observe that x=4 works, plug into second expression. Done!
Ahahahaha! Nice 🤪
Beautiful! Thank you.
Thank you too!
2:58 When you had xsqrtx+2sqrtx=x^2-4. sqrtx(x+2)=(x+2)(x-2) x>0 so ➗️ x+2 both sides
sqrtx=x-2 then sq both sides you get x=1 and x=4. x=1 doesn't work x=4 is the only solution.
Does anyone else feel that if the difference between two squares is not used in a few days, it has to be part of the current solution?
√x + 2/√x = x - 4/x
Because the left-hand side is always positive, the right-hand side should be too…
Squaring both sides yields an extra condition for real solutions, namely that x - 4/x >= 0.
If we solve this inequality, we get x² >= 4 which results in -2>=x and x>=2.
Because of the first condition (x>=0) the final quadrature condition becomes x >= 2
(x-1)(x-4)(x+2)²=0
This way there’s only one real root and that’s 4.
Why is the second part of the problem even there if we're going to solve for x in every method?
How about expanding to the quartic, and then creating two quadratics: x^2-5x-a, and x^2+bx+c. Multiply them together and set the coefficients of each power of x to the ones in the quartic. b and c are easy to solve for (you don't even need c) and then a is your answer. Skip solving for x.
By inspection x = 4 works. So 16 - 20 = -4.
Syber math, you are so funny, why your 1st method usually 3rd :)
Why not? 😜
-4
Pozdrawiam serdecznie i życzę miłego dnia
-4.
I used the second method. But I didn't understand why you want to evaluate x^2-5x instead of simply solving the equation for x. Probably to make it more tricky?😁
Nice. My initial thought was that I could get it directly without finding x but then I forgot what I was thinking 🤪
See ralf’s comment
@@SyberMath you can also solve it without solve x with your 3 method: when you got x-2=sqrtx ,by x^2-5x=(x-2)^2-x-4=x-x-4=-4
@@SyberMath I've just seen it. Fantastic! 👍
Try to find common factor. sqrt(x)+2'sqrt(x)=x-4/x, sqrt(x)+2/sqrt(x)=(sqrt(x)+2/sqrt(x))(sqrt(x)-2/sqrt(x))). So sqrt(x)+2/sqrt(x)=0 or 1=sqrt(x)-2/sqrt(x). x=4
My own solution: √x + 4/x = x - 2/√x;
√x + 2/√x = x - 4/x;
(√x + 2/√x)(√x - 2/√x) = x - 4/x = √x + 2/√x ≠0;
√x - 2/√x = 1 (√x = y)
y² - y - 2 = 0; y = -1, 2; but √n >= 0, √x = 2, x = 4.
4² -5*4 = -4
I have solved with second method
Answer =-4
Variation on #2:
Multiply both sides by x:
x√x + 4 = x² - 2√x →
√x(x + 2) = x² - 4 →
√x = x - 2 (÷ both sides by the nonzero quantity x+2) →
x = x² - 4x + 4 →
x² - 5x = -4 ✅
Great!
x = 4; x²-5x=-4
Why?
sqrt(x) + 4/x = x - 2/sqrt(x) x•sqrt(x) + 4 = x^2 - 2•sqrt(x) x•sqrt(x) + 2•sqrt(x) = sqrt(x)•(x + 2) = x^2 - 4 = (x - 2)•(x + 2) x + 2 = 0 or sqrt(x) = x - 2 x = -2 or x - sqrt(x) - 2 = sqrt(x)^2 - sqrt(x) - 2 = (sqrt(x) - 2)•(sqrt(x) + 1) = 0 FALSE or sqrt(x) - 2 = 0 x = 4.
x = 4 ==> x^2 - 5•x = 4^2 - 5•4 = 16 - 20 = -4.
The comments have been saying that x = -2 is possible if complex numbers are allowed. However, this is inaccurate. The radical symbol is not well-defined for complex numbers. It is only well-defined for ordered fields which are square-root closed.
The task has been formulated unfairly. Your formulation assumes : Find x^2-5x without finding x.
Not really
@@sand7861 normally to avoid the confusion in the official tasks they state :solve.... and write the answer in form....
who says you cannot find x, then the solution being asked for??
@@ytlongbeach it said : find x^2-5x