yeah well, it only works the this certain type of problem. If you watch blackpenredpen’s video on the dear tejas challenge, I think it’s called quadratic in terms of 5, you can see another problem like this where it works, but it doesn’t work normally.
@@leif1075 kind of, but it’s quite impossible to solve this equation otherwise, so solving this question would be something no one would think of anyways. So idk. Other methods that I’ve seen need very high levels of thinking. This way is just remember this type of question
@@koennako2195 no its not impossible..he even uses a second methid..thiugh substitution of this type is also funky and not something most ppl would.think..But surely there is another way.
3rd method: Solve a quartic equation using Ferrari's method of completing the square: x^2 - 4 = sqrt(4 - x) (x^2 - 4)^2 = 4 - x x^4 - 8x^2 + 16 = 4 - 1x x^4 - 8x^2 + 1x + 12 = 0 (Reduced or depressed quartic equation). Isolate the 4th power: x^4 = 8x^2 - 1x - 12 Multiply by 4, in order to avoid fractions: 4x^4 = 32x^2 - 4x - 48 Complete the square by adding 4zx^2 + z^2 to both sides: 4x^4 + 4zx^2 + z^2 = (4z + 32)x^2 - 4x + (z^2 - 48) Rewrite the left side as a perfect square: (2x^2 + z)^2 = 4(z + 8)x^2 - 4x + (z^2 - 48) Condition for the right side being a perfect square, too: ax^2 + bx + c is a perfect square iff. discriminant D = b^2 - 4ac = 0, i.e. iff. b^2 = 4ac. In this case: (-4)^2 = 4 * 4(z + 8)(z^2 - 48) 16 = 16(z + 8)(z^2 - 48) 1 = (z + 8)(z^2 - 48) 1 = z^3 + 8z^2 - 48z - 384 0 = z^3 + 8z^2 - 48z - 385 This cubic resolvent has three real solutions. I take the only integer solution z = -7 and plug it into the quartic equation: (2x^2 - 7)^2 = 4*(-7 + 8)x^2 - 4x + ((-7)^2 - 48) (2x^2 - 7)^2 = 4*1*x^2 - 4x + (49 - 48) (2x^2 - 7)^2 = 4x^2 - 4x + 1 And now we can rewrite the right side as a perfect square, too: (2x^2 - 7)^2 = (2x - 1)^2 Taking the square root on both sides (A^2 = B^2 => A = +-B): 2x^2 - 7 = +-(2x - 1) So we get two quadratic equations: 2x^2 - 7 = 2x - 1 2x^2 - 2x - 6 = 0 x^2 - x - 3 = 0 and 2x^2 - 7 = -2x + 1 2x^2 + 2x - 8 = 0 x^2 + x - 4 = 0 The solutions of these are x1, x2 = (1 +- sqrt(13))/2 x3, x4 = (-1 +- sqrt(17))/2 the same as in the video. But I currently don't yet understand how to check which solutions are valid and which are false positives, except by putting approximate the values into the original equation.
@@SyberMath Yes, but being comprised in the defintion set (is "defintion set" the correct term in English?) is a necessary but no sufficient condition for the validity of a solution candidate. For example, the equation sqrt(x) = -3 has the definition set R+0 (all real numbers >= 0). If I square the equation, I get x = 9. The 9 is in the definition set but no valid solution. So I have to check every solution by plugging it into the original equation. I can of course check the approximate values this way, or check where the two graphs of y = sqrt(4 - x) and y = x^2 - 4 intersect.
As in method 1, after squaring both sides you get x^4 - 8x^2 +x +12. since there is no cubic term I guessed the quartic was the product of two quadratics: (x^2 +Bx +C)(X^2 - Bx +E). Then comparing coefficients you get constant = C*E =12 coeff of x = B*(E -C)= 1 coeff of x^2 = (C + E - B^2)= -8. This system is satisfied if B=1 C= -4 and E= -3. Pluging in these values you get the quadratics in method 1.
That is what I did, but I had to guess to get the values of B, C, and E. The guess was helped by B*(E-C) = 1: If B is an integer it divides 1 so I might as well guess B=1. Then E-C=1 and C*E=12 leading quickly to C=-4 and E=-3, or C=3 and E=4 which leads to the same factorization.
@@rorydaulton6858 Sometimes the method of undetermined coefficients can lead to some pretty crazy systems. In this case the system was fairly simple and the G&C method (guess and check) was not that hard to implement.
@@SyberMath why in God's name would anyone think of thst..even Ramanujan wouldn't thinknof thst why not just solve by breaking it into two quadratics..(x^2 + bx + c)(x^2 + dx + f)...THAT should work..isn't this just crazy??
The domain is xor equal to 0 since the sqrt(4-x) is always nonnegative. Which means |x|>or equal to 2. Or in other words x< or equal to -2 or x> or equal to 2. The two extraneous roots come from the negative sqrt(4-x) which we don’t consider by definition.
y=Sqrt(4-x) and y=x^2 - 4 are same shape parabolas rotated around coordinate origin and they are symmetrical for y=-x line. This allows to find roots on this symmetry line with new equation sqrt(4-x)=-x. Also this equation gives us quadratic polynomial which can be used to divide original 4 degree polynomial and get the second factor.
@@SyberMath I see the graph but how exactly would you divine the second factor ? Don't you agree this substitution you did here is infuriating because nonone wpuldmever think to solve for 4 since normally you eliminate the variables not the constants..didnt it take you a while to see that..and surely there is another method more intuitive or more clearly logical than these two ways shown here?
sqrt(4-x)=x²-4 belongs to a special equation sqrt(y-x)=x²-y which upon squaring we get y-x=x⁴-2xy+y² We may consiider it as quartic equation in x, or as quadratic one in y. As quartic equation in x, we can obtain the solution trough Ferrri's method: x as function of y. As quadratic equation, the solution is easier to obtain. There are only two roots: y as function of x --> x is then the inverse function of y.
I really enjoyed!!!! I did not have any clue of solving it....but then I saw the title "Olympiad Maths" so I said to myself "you have some small excuse for not to...."
How about dividing both sides by (4-x)? Then we have: 1/sqrt(4-x)=-x-4 Square both sides: 1/(4-x)=x^2+8x+16 *(x-4) x^3+4x^2-16x-64=-1 That way we easily get a cubic equation.
Hi Syber, thank you for daily uploading! I am going take part in Republican Olympiad in Moldova this week, so i'm exercising a lot. Could you please help me with this exercise: x and y are real numbers that verify this equality x^2 + y^2 - 2x +12y + 33 = 0. Proof that x > y Thank you a lot!
@@monishrules6580 complete the squares and rewrite in the form of the equation of a circle (1) (x-1)^2 + (y+6)^2 = 2^2. Two double inequalities follow: (2) -2
sqrt(4-x)=x^2-4]=[ 4-x= 4/x= 4x-16]=[4/x=4x/16= 4/x= x/4]=[ 4x-4x=1] = 4x/4x=4x-1=4x/1=4x=x^4=4 x=1 AcademiC Uppsala Universty 23 september nobel prize Fields Prize Abel Prize Medal diplom
I would never ever think of letting a = 4 and transform the equation into a quadratic in a! Genius.
yeah well, it only works the this certain type of problem. If you watch blackpenredpen’s video on the dear tejas challenge, I think it’s called quadratic in terms of 5, you can see another problem like this where it works, but it doesn’t work normally.
@@koennako2195 isn't thst then stupid and infuriating. Why do something NO 0NE would EVER think of..
@@leif1075 kind of, but it’s quite impossible to solve this equation otherwise, so solving this question would be something no one would think of anyways. So idk. Other methods that I’ve seen need very high levels of thinking. This way is just remember this type of question
@@koennako2195 no its not impossible..he even uses a second methid..thiugh substitution of this type is also funky and not something most ppl would.think..But surely there is another way.
3rd method:
Solve a quartic equation using Ferrari's method of completing the square:
x^2 - 4 = sqrt(4 - x)
(x^2 - 4)^2 = 4 - x
x^4 - 8x^2 + 16 = 4 - 1x
x^4 - 8x^2 + 1x + 12 = 0
(Reduced or depressed quartic equation).
Isolate the 4th power:
x^4 = 8x^2 - 1x - 12
Multiply by 4, in order to avoid fractions:
4x^4 = 32x^2 - 4x - 48
Complete the square by adding 4zx^2 + z^2 to both sides:
4x^4 + 4zx^2 + z^2 = (4z + 32)x^2 - 4x + (z^2 - 48)
Rewrite the left side as a perfect square:
(2x^2 + z)^2 = 4(z + 8)x^2 - 4x + (z^2 - 48)
Condition for the right side being a perfect square, too:
ax^2 + bx + c is a perfect square iff. discriminant D = b^2 - 4ac = 0, i.e. iff. b^2 = 4ac. In this case:
(-4)^2 = 4 * 4(z + 8)(z^2 - 48)
16 = 16(z + 8)(z^2 - 48)
1 = (z + 8)(z^2 - 48)
1 = z^3 + 8z^2 - 48z - 384
0 = z^3 + 8z^2 - 48z - 385
This cubic resolvent has three real solutions. I take the only integer solution z = -7 and plug it into the quartic equation:
(2x^2 - 7)^2 = 4*(-7 + 8)x^2 - 4x + ((-7)^2 - 48)
(2x^2 - 7)^2 = 4*1*x^2 - 4x + (49 - 48)
(2x^2 - 7)^2 = 4x^2 - 4x + 1
And now we can rewrite the right side as a perfect square, too:
(2x^2 - 7)^2 = (2x - 1)^2
Taking the square root on both sides (A^2 = B^2 => A = +-B):
2x^2 - 7 = +-(2x - 1)
So we get two quadratic equations:
2x^2 - 7 = 2x - 1
2x^2 - 2x - 6 = 0
x^2 - x - 3 = 0
and
2x^2 - 7 = -2x + 1
2x^2 + 2x - 8 = 0
x^2 + x - 4 = 0
The solutions of these are
x1, x2 = (1 +- sqrt(13))/2
x3, x4 = (-1 +- sqrt(17))/2
the same as in the video. But I currently don't yet understand how to check which solutions are valid and which are false positives, except by putting approximate the values into the original equation.
Very nice! You can check the domain of the original equation
@@SyberMath Yes, but being comprised in the defintion set (is "defintion set" the correct term in English?) is a necessary but no sufficient condition for the validity of a solution candidate. For example, the equation sqrt(x) = -3 has the definition set R+0 (all real numbers >= 0). If I square the equation, I get x = 9. The 9 is in the definition set but no valid solution. So I have to check every solution by plugging it into the original equation. I can of course check the approximate values this way, or check where the two graphs of y = sqrt(4 - x) and y = x^2 - 4 intersect.
As in method 1, after squaring both sides you get x^4 - 8x^2 +x +12. since there is no cubic term I guessed the quartic was the product of two quadratics:
(x^2 +Bx +C)(X^2 - Bx +E). Then comparing coefficients you get constant = C*E =12 coeff of x = B*(E -C)= 1 coeff of x^2 = (C + E - B^2)= -8.
This system is satisfied if B=1 C= -4 and E= -3. Pluging in these values you get the quadratics in method 1.
That is what I did, but I had to guess to get the values of B, C, and E. The guess was helped by B*(E-C) = 1: If B is an integer it divides 1 so I might as well guess B=1. Then E-C=1 and C*E=12 leading quickly to C=-4 and E=-3, or C=3 and E=4 which leads to the same factorization.
@@rorydaulton6858 Sometimes the method of undetermined coefficients can lead to some pretty crazy systems. In this case the system was fairly simple and the
G&C method (guess and check) was not that hard to implement.
Nice!
The system of equations made me scream in awe
😍
@@SyberMath why in God's name would anyone think of thst..even Ramanujan wouldn't thinknof thst why not just solve by breaking it into two quadratics..(x^2 + bx + c)(x^2 + dx + f)...THAT should work..isn't this just crazy??
The domain is xor equal to 0 since the sqrt(4-x) is always nonnegative. Which means |x|>or equal to 2. Or in other words x< or equal to -2 or x> or equal to 2. The two extraneous roots come from the negative sqrt(4-x) which we don’t consider by definition.
Also factors to (x^2 - 7/4)^2 = (x - 1/2)^2 with same results. Take x^4 - 8x^2 + x + 12 = 0 break into
x^4 - 7x^2 - 1x^2 + x + 49/4 - 1/4 = x^4 - 7x^2 + 49/4 + (- 1x^2 + x- 1/4) = 0
위 식을 양쪽을 제곱하면 4차식이 나오고, x
Bro, use english.
@@SarthakJoshi_maths -- * *English*
@@robertveith6383Ching chong ching xang-a-lang xong xong xi piao piao bing chiling -420 social credit rong xie zung
y=Sqrt(4-x) and y=x^2 - 4 are same shape parabolas rotated around coordinate origin and they are symmetrical for y=-x line. This allows to find roots on this symmetry line with new equation sqrt(4-x)=-x. Also this equation gives us quadratic polynomial which can be used to divide original 4 degree polynomial and get the second factor.
That's interesting!
nice!
@@SyberMath I see the graph but how exactly would you divine the second factor ? Don't you agree this substitution you did here is infuriating because nonone wpuldmever think to solve for 4 since normally you eliminate the variables not the constants..didnt it take you a while to see that..and surely there is another method more intuitive or more clearly logical than these two ways shown here?
Marvelous👌🏻
Thank you!
Very interesting!!! ❤❤❤
Glad you think so! 🥰
Both the methods are amazing! Just wow! 👏👏👏❤️
Thanks so much 😊
@@SyberMath 😇🙏
2nd method just left me dumbfounded.
Great solution!
Glad it helped!
God of substitution
sqrt(4-x)=x²-4 belongs to a special equation sqrt(y-x)=x²-y which upon squaring we get
y-x=x⁴-2xy+y²
We may consiider it as quartic equation in x, or as quadratic one in y. As quartic equation in x, we can obtain the solution trough Ferrri's method: x as function of y. As quadratic equation, the solution is easier to obtain. There are only two roots: y as function of x --> x is then the inverse function of y.
Very nice! I liked the very quick checking solutions at 7:35 😂💯
I did not get this part
Glad you liked it!
👍😉
A very smart solution. I just squared everything and then got a quartic equation and solved it with ferrari's method.
I really enjoyed!!!! I did not have any clue of solving it....but then I saw the title "Olympiad Maths" so I said to myself "you have some small excuse for not to...."
√(4-x) =x^2-4
4+√(4-x) =x^2
√(4-x)(√(4-x) +1) =x(x-1)
√(4-x) =B
X=A
Solve for either a or B from the quadratic and set x equal to it
How about dividing both sides by (4-x)?
Then we have:
1/sqrt(4-x)=-x-4
Square both sides:
1/(4-x)=x^2+8x+16
*(x-4)
x^3+4x^2-16x-64=-1
That way we easily get a cubic equation.
x^2 - 4 divided by 4 - x is not equal to -x - 4, though.
Awesome, many thanks, Sir, you are great! 🙂
So nice of you
2nd solution is much better
Yes,but the first method is really cool!
Good one.
Thanks!
x=(1+√13)/2, -(1+√17)/2
This might take some time.=)
hehe 😜
Very interesting method.👍
Glad you think so!
Elevo al quadrato e ottengo x^4-8x^2+x+12=0...calcolo e risolvo la quartica in (x^2-7/2)^2-(x-1/2)^2=0...
that is simply B..E..A..U..TIFUL
Thank you!
I enjoyed it. Thank you. (You asked us to let you know.)
Glad you enjoyed it! 🤩
Hi Syber, thank you for daily uploading! I am going take part in Republican Olympiad in Moldova this week, so i'm exercising a lot. Could you please help me with this exercise:
x and y are real numbers that verify this equality x^2 + y^2 - 2x +12y + 33 = 0. Proof that x > y
Thank you a lot!
Nevermind, just solved it 😃
Post solution pls
Nice! Good luck
@@monishrules6580 complete the squares and rewrite in the form of the equation of a circle (1) (x-1)^2 + (y+6)^2 = 2^2. Two double inequalities follow: (2) -2
@@JordHaj 😊
I did it a very different way. Turns out I just made a big mess and gave up cause it was all wrong.
Never give up!
sqrt(4-x)=x^2-4]=[ 4-x= 4/x= 4x-16]=[4/x=4x/16= 4/x= x/4]=[ 4x-4x=1] = 4x/4x=4x-1=4x/1=4x=x^4=4 x=1
AcademiC Uppsala Universty 23 september nobel prize Fields Prize Abel Prize Medal diplom
2 stay or 2 unsubscribe. That is the ℝ question.
4 ℝ's? 😜😄
I mean, it would be 4-x = (x-2)(x+2), no + or -, lol...I mean...
[(x-2)(x+2)]^2, that is...
...a question like this on YOutube, there is probably a trick to get out of solving a quartic, lol...
Only fools rush in! Square both sides, 4 -> 3 -> 2 vieta ... 😃
Nice
Thanks