This is from the Regional Math Olympiad. The trick here is that not all polynomials are solved in terms of x. Sometimes the other unknown variable becomes the key.
I solved this using a more traditional approach, the method of undetermined coefficients to factor a quartic. We have x⁴ − 2ax² − x + (a² − a) = 0 There is no term with x³, so if this is to factor into two quadratics we must have (x² + px + q₁)(x² − px + q₂) = 0 for some p, q₁, q₂. Expanding this we have x⁴ + (q₁ + q₂ − p²)x² + p(q₂ − q₁)x + q₁q₂ = 0 and equating corresponding coefficients we have q₁ + q₂ − p² = −2a p(q₂ − q₁) = −1 q₁q₂ = a(a − 1) From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1 and then it is not hard to find the solution triple (p, q₁, q₂) = (1, −(a − 1), −a) as well as (p, q₁, q₂) = (−1, −a, −(a − 1)) because inverting the sign of p will simply swap the values of q₁ and q₂ for the same factorization. With (p, q₁, q₂) = (1, −(a − 1), −a) we have (x² + x − a + 1)(x² − x − a) = 0 so x² + x − a + 1 = 0 ∨ x² − x − a = 0 and then it is just a matter of finding the values of a for which both of these quadratics have nonnegative discriminants. So we have 1 + 4(a − 1) ≥ 0 ⋀ 1 + 4a ≥ 0 which gives a ≥ ¾
@@PrimeNewtons Yet another way to factor this quartic into two quadratics would be as follows. x⁴ − 2ax² − x + a² − a = 0 (x² − a)² − x − a = 0 (x² − a)² + (x² − a) − x² − x = 0 (x² − a)² + (x² − a) − (x² + x) = 0 (x² − a)² + (x² − a) + ¼ − (x² + x + ¼) = 0 (x² − a + ½)² − (x + ½)² = 0 (x² − a + ½ + x + ½)(x² − a + ½ − x − ½) = 0 (x² + x − a + 1)(x² − x − a) = 0
I understand your approach, but don't understand the assumption about why it must be able to be factored. Not all polynomials with real solutions can be factorised.
@@Straight_Talk Actually any polynomial with real coefficients can be factored into linear and quadratic factors with real coefficients. This is a consequence of the fundamental theorem of algebra (which guarantees the existence of at least a single zero of any polynomial), the polynomial factor theorem, and the fact that complex zeros of polynomials with real coefficients always occur as conjugate pairs. So, any quartic polynomial with real coefficients can be factored into two quadratics with real coefficients, regardless of whether or not this quartic polynomial has real zeros. Of course, we are not talking here of factoring polynomials with _integer_ coefficients into linear or quadratic factors with _integer_ coefficients. As is well known, a quadratic polynomial ax² + bx + c with integer coefficients a, b, c (a ≠ 0) can be factored into linear factors with integer coefficients _if and only if_ the discriminant b² − 4ac of this quadratic is the square of an integer.
I also got to your three system of equations for p.q's ~ a, but where you claim _'From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1'_ to me that is not so obvious a scenario, I mean they could be fractional?
I recognized this problem immediately: Edit: (either use nested radical or solve as quadratic in terms of a!) x^4 - 2ax^2 + x + a^2 - a = 0 x^4 -2ax^2 + a^2 = a - x (a - x^2)^2 = a-x 1st) a - x^2 = +sqrt(a-x) Solve for x: x = sqrt(a - sqrt(a- x)) x= sqrt(a - x) x^2 = a - x x^2 + x - a = 0 x = (-1+-sqrt(4a + 1))/2 2nd) a - x^2 = (-1)sqrt(a- x) Too lazy for this, so did long division of original polynomial (x^4 -2ax^2 + x + a^2 - a)/(x^2 + x - a) = x^2 - x + 1 - a x= (1+-sqrt(4a-3))/2 Then solving for real domain of a: 4a-3 >= 0 a >= 3/4 4a + 1 >= 0 a >= -1/4 1st restriction is more descriptive, so *a>=3/4*
I solved this problem using derivation. We know that the graph of f(x) needs to intersect the x-axis at 4 points for there to be 4 real solutions. Between each of these points, there is a local minimum or maximum, which means that the derivative of the function has to intersect the x-axis at at least 2 points. To find out under which conditions this is true, we find an expression for the x-value of the local minimum of f'(x) by setting the second derivative to 0: f''(x)=0. This yields an x-value of sqrt(a/3). Because the local minimum of the first derivative has to lie on or below the x-axis, we can solve f'(sqrt(a/3))3/4
I don't think your reasoning is correct. If the graph of the quartic intersects the x-axis at 4 distinct points (meaning we have 4 distinct real zeros) then the first derivative has _three_ distinct zeros, one in between each two consecutive zeros of the quartic. In fact, this gives yet another way to approach this problem. The condition for the first derivative to be zero is 4x³ − 4ax − 1 = 0 or x³ − ax − ¼ = 0 This is a depressed cubic x³ + px + q = 0 which has three distinct real zeros if and only if (½q)² + (⅓p)³ < 0 so a must then satisfy (−⅛)² + (−⅓a)³ < 0 which indeed gives a > ¾. This is a _necessary_ condition for the quartic x⁴ − 2ax² − x + (a² − a) to have four distinct real zeros but we would still need to prove that this is also a _sufficient_ condition. In fact we must prove that the two local minima of x⁴ − 2ax² − x + (a² − a) are both negative _and_ that its local maximum is positive for a > ¾ to have four distinct real zeros and I don't see you doing that. The two zeros of the second derivative 12x² − 4a give the positions of the inflection points at x = √(a/3) and x = −√(a/3) but your claim that the point where the first derivative reaches a local minimum, that is, at x = √(a/3), has to lie on or below the x-axis makes no sense. In fact, both inflection points can lie above the x-axis and then the graph of the quartic can still cross the x-axis at four distinct points. And in fact if we substitute x = √(a/3) in f'(x) = 4x³ − 4ax − 1 and solve for f'(√(a/3)) < 0 we do _not_ get a > ¾.
@@NadiehFan Thank you for your observations. First, I agree with your point about the first derivative needing to have at least 3 zeroes for the function to have 4 distinct real roots. The reason I said there need to be at least two roots, is because I wanted to include identical roots. Your other point is more interesting, because I wrongly assumed the two local minima of the function to always be negative, which seems to be true for this particular function, but not for a general quartic. In a general case, one could remove all the real roots by adding a constant to the function, which would not be visible from the derivative. My question then becomes, what is it about the particular constant of a^2-a that ensures that the local minima are negative?
I think the solutions are different. Should be a=x^2-x and x^2-1. Graphing the two functions of a on the same graph we got 2 quadratics one greater than -1 and the other always greater than -1/4. So if I intersect the grap with a horizontal line to get 4 intersections then a must be greater than -1/4
Thanks #primenewtons for this interesting solving method. It is interesting to point that for a=3/4 the equation has got only two different roots which one is -3/2 as a single root and the other is 1/2 which is a triple root (being the simple second root of the first factor and at the same time the double root of the second factor). As well, it can be easily seen and graphed with the use of Desmos which allows to add "a" as a cursor value showing that the real root values number is as follow: -0 for a
You have no need to use the quadratic equation solution. For this task, it is sufficient and necessary to consider the value of the discriminant. And when solving a quadratic equation, solving the discriminant solves the question of the nature of their solutions. Don't stop learning. I was born in 1950 and I'm still learning.😎
a^2-(2x^2+1)a+x(x^3+1)=0 a=(2x^2+1+-sqrt(4x^4+4x^2+1-4x^4-4x))/2 a=(2x^2+1+-sqrt(4x^2-4x+1))/2 a=(2x^2+1+-(2x-1))/2 a1=x^2+x a2=x^2-x+1 x^2+x-a1=0 x1=(-1+-sqrt(1+4a1))/2 x^2-x+(1-a2)=0 x2=(1+-sqrt(4a2-3))/2 1+4a1>=0; 4a2-3>=0 a1>=-1/4; a2>=3/4 since both must be true for all 4 roots to be real, a>=3/4
That's an excellent question, you are referring of course to the i-th derivative operator Dⁱ (f(x)) = fⁱ (x). From _D-operator theory_ , F(D)e^(ax) ≡ e^(ax). F(a) so, for imaginary indices operation; Dⁱ [e^(ax)] = e^(ax). aⁱ and for polynomial functions, Dⁱ [xⁿ] = α(n)xⁿ-ⁱ where function α(n) is complex. there are similar formulae for trig functions. This is the basis for Laplace operators in solving differential equations, so not so esoteric.
Interesante, yo para evitarme esas cosas, simplemente aplique aspa doble especial a la ecuacion de 4to grado, me dejaron 2 expresiones de 2do grado, discriminante >=0 para ambos casos con la 'a' y finalmente intersectar
@@shj_happy10 no i know this but more i know If under the square root there is a number squared, then the number under the modulus comes out of it because 3 square is 9 and -3 square is also 9 and √(-3)^2 =3 not -3 i don't know you understand what i explained because my English not fluently :)
@@shj_happy10noo bro she is actually correct √x² = |x| Because u know that's how modulus of a square root function is defined in order not to break fundamental theorem of zeroes of polynomials So √9 is only 3, not -3 U may see it like that √9 = √3² = |3| = 3 √9 = √(-3)² = |-3| = 3 So as per the definition we are having one answer and that's correct
ok although it seems counterproductive for a representative of math and logic to throw in that non-sequitor at the end about loving darkness due to evil deeds.
The answer doesn’t make sense. One of the conditions for a is that it has to be greater than or equal to -1/4. That should be the correct answer. If we say that the answer is a is greater than or equal to 3/4, then that eliminates -1/4 because 3/4 is greater than -1/4. But we’ve already established that a is greater than or equal to -1/4 is a condition for a. Therefore, the correct answer to the question is this: a is greater than or equal to -1/4.
If you check a = 0, you will only get two real roots (the other two are non-real), so that eliminates the condition that a is greater than or equal to -1/4. Hope this helps!
![Find (x+y+z) [Harvard-MIT] Guts contest](http://i.ytimg.com/vi/Pe0s5N_jkLA/mqdefault.jpg)
I too am unclear why you think that a>-1/4 AND a>3/4. You might be correct, but can you please explain why it's not a.-1/4 OR a>3/4?
When a variable satisfies two inequalities, it satisfies the intersection.
@@PrimeNewtons when a ≥ 3/4, there are 4 real roots, when a ≥ -1/4 then 2 roots.
I solved this using a more traditional approach, the method of undetermined coefficients to factor a quartic. We have
x⁴ − 2ax² − x + (a² − a) = 0
There is no term with x³, so if this is to factor into two quadratics we must have
(x² + px + q₁)(x² − px + q₂) = 0
for some p, q₁, q₂. Expanding this we have
x⁴ + (q₁ + q₂ − p²)x² + p(q₂ − q₁)x + q₁q₂ = 0
and equating corresponding coefficients we have
q₁ + q₂ − p² = −2a
p(q₂ − q₁) = −1
q₁q₂ = a(a − 1)
From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1 and then it is not hard to find the solution triple (p, q₁, q₂) = (1, −(a − 1), −a) as well as (p, q₁, q₂) = (−1, −a, −(a − 1)) because inverting the sign of p will simply swap the values of q₁ and q₂ for the same factorization. With (p, q₁, q₂) = (1, −(a − 1), −a) we have
(x² + x − a + 1)(x² − x − a) = 0
so
x² + x − a + 1 = 0 ∨ x² − x − a = 0
and then it is just a matter of finding the values of a for which both of these quadratics have nonnegative discriminants. So we have
1 + 4(a − 1) ≥ 0 ⋀ 1 + 4a ≥ 0
which gives
a ≥ ¾
I thought of that, too. Nice approach.
@@PrimeNewtons Yet another way to factor this quartic into two quadratics would be as follows.
x⁴ − 2ax² − x + a² − a = 0
(x² − a)² − x − a = 0
(x² − a)² + (x² − a) − x² − x = 0
(x² − a)² + (x² − a) − (x² + x) = 0
(x² − a)² + (x² − a) + ¼ − (x² + x + ¼) = 0
(x² − a + ½)² − (x + ½)² = 0
(x² − a + ½ + x + ½)(x² − a + ½ − x − ½) = 0
(x² + x − a + 1)(x² − x − a) = 0
I understand your approach, but don't understand the assumption about why it must be able to be factored.
Not all polynomials with real solutions can be factorised.
@@Straight_Talk Actually any polynomial with real coefficients can be factored into linear and quadratic factors with real coefficients. This is a consequence of the fundamental theorem of algebra (which guarantees the existence of at least a single zero of any polynomial), the polynomial factor theorem, and the fact that complex zeros of polynomials with real coefficients always occur as conjugate pairs.
So, any quartic polynomial with real coefficients can be factored into two quadratics with real coefficients, regardless of whether or not this quartic polynomial has real zeros.
Of course, we are not talking here of factoring polynomials with _integer_ coefficients into linear or quadratic factors with _integer_ coefficients. As is well known, a quadratic polynomial ax² + bx + c with integer coefficients a, b, c (a ≠ 0) can be factored into linear factors with integer coefficients _if and only if_ the discriminant b² − 4ac of this quadratic is the square of an integer.
I also got to your three system of equations for p.q's ~ a, but
where you claim
_'From p(q₂ − q₁) = −1 we may already suspect that we have either p = 1 and q₂ − q₁ = −1 or p = −1 and q₂ − q₁ = 1'_
to me that is not so obvious a scenario, I mean they could be fractional?
Your last statement a ≥ 3/4 is not correct.
when a ≥ 3/4, there are 4 real roots, when a ≥ -1/4 then 2 roots.
I recognized this problem immediately:
Edit: (either use nested radical or solve as quadratic in terms of a!)
x^4 - 2ax^2 + x + a^2 - a = 0
x^4 -2ax^2 + a^2 = a - x
(a - x^2)^2 = a-x
1st) a - x^2 = +sqrt(a-x)
Solve for x:
x = sqrt(a - sqrt(a- x))
x= sqrt(a - x)
x^2 = a - x
x^2 + x - a = 0
x = (-1+-sqrt(4a + 1))/2
2nd) a - x^2 = (-1)sqrt(a- x)
Too lazy for this, so did long division of original polynomial
(x^4 -2ax^2 + x + a^2 - a)/(x^2 + x - a)
= x^2 - x + 1 - a
x= (1+-sqrt(4a-3))/2
Then solving for real domain of a:
4a-3 >= 0
a >= 3/4
4a + 1 >= 0
a >= -1/4
1st restriction is more descriptive, so
*a>=3/4*
This is probably the shortest way and...on my opinion the more elegant. Bravo! (Iam French).
I am from Bangladesh
'Too lazy for this,..' , but not too lazy to work it all out. 👌
I think solving as a quadratic in terms of a is a little quicker as it gives both quadratics pretty quickly.
Surely the answer is a>= -1/4. Values of a between -1/4 and +3/4 will give two real roots of the equation (and two imaginary roots).
the question is "all real roots", so can't have 2 imaginary roots
@@TheThendria I see. The thumbnail is different. It just asks for "real roots".
Yes, I was confused about that as well
No,it says "has real roots",0:00 which means has at least one real root
So the answer should be -1/4 which makes the equation possess at least one root
I solved this problem using derivation. We know that the graph of f(x) needs to intersect the x-axis at 4 points for there to be 4 real solutions. Between each of these points, there is a local minimum or maximum, which means that the derivative of the function has to intersect the x-axis at at least 2 points. To find out under which conditions this is true, we find an expression for the x-value of the local minimum of f'(x) by setting the second derivative to 0: f''(x)=0. This yields an x-value of sqrt(a/3). Because the local minimum of the first derivative has to lie on or below the x-axis, we can solve f'(sqrt(a/3))3/4
I don't think your reasoning is correct. If the graph of the quartic intersects the x-axis at 4 distinct points (meaning we have 4 distinct real zeros) then the first derivative has _three_ distinct zeros, one in between each two consecutive zeros of the quartic.
In fact, this gives yet another way to approach this problem. The condition for the first derivative to be zero is
4x³ − 4ax − 1 = 0
or
x³ − ax − ¼ = 0
This is a depressed cubic x³ + px + q = 0 which has three distinct real zeros if and only if (½q)² + (⅓p)³ < 0 so a must then satisfy
(−⅛)² + (−⅓a)³ < 0
which indeed gives a > ¾.
This is a _necessary_ condition for the quartic x⁴ − 2ax² − x + (a² − a) to have four distinct real zeros but we would still need to prove that this is also a _sufficient_ condition.
In fact we must prove that the two local minima of x⁴ − 2ax² − x + (a² − a) are both negative _and_ that its local maximum is positive for a > ¾ to have four distinct real zeros and I don't see you doing that. The two zeros of the second derivative 12x² − 4a give the positions of the inflection points at x = √(a/3) and x = −√(a/3) but your claim that the point where the first derivative reaches a local minimum, that is, at x = √(a/3), has to lie on or below the x-axis makes no sense. In fact, both inflection points can lie above the x-axis and then the graph of the quartic can still cross the x-axis at four distinct points. And in fact if we substitute x = √(a/3) in f'(x) = 4x³ − 4ax − 1 and solve for f'(√(a/3)) < 0 we do _not_ get a > ¾.
@@NadiehFan Thank you for your observations. First, I agree with your point about the first derivative needing to have at least 3 zeroes for the function to have 4 distinct real roots. The reason I said there need to be at least two roots, is because I wanted to include identical roots.
Your other point is more interesting, because I wrongly assumed the two local minima of the function to always be negative, which seems to be true for this particular function, but not for a general quartic. In a general case, one could remove all the real roots by adding a constant to the function, which would not be visible from the derivative.
My question then becomes, what is it about the particular constant of a^2-a that ensures that the local minima are negative?
@@NadiehFanYour approach is very interesting, and it does solve the problem.
2:53 a^2 - (2x^2 + 1)a + (x^4 + x) = 0
since x^4 + x = x(x^3 +1) = x(x + 1)(x^2 - x + 1)
we can factorize a^2 - (2x^2 + 1)a + (x^4 + x) = ( a - (x^2 + x) )( a - (x^2 - x + 1) )
I think the solutions are different. Should be a=x^2-x and x^2-1. Graphing the two functions of a on the same graph we got 2 quadratics one greater than -1 and the other always greater than -1/4. So if I intersect the grap with a horizontal line to get 4 intersections then a must be greater than -1/4
notice that sqrt((2x-1)^2) = |2x-1|,in this problem there is a ± before sqrt((2x-1)^2) ，so it can be wrote as ±（2x-1）
Thanks #primenewtons for this interesting solving method. It is interesting to point that for a=3/4 the equation has got only two different roots which one is -3/2 as a single root and the other is 1/2 which is a triple root (being the simple second root of the first factor and at the same time the double root of the second factor).
As well, it can be easily seen and graphed with the use of Desmos which allows to add "a" as a cursor value showing that the real root values number is as follow:
-0 for a
You have no need to use the quadratic equation solution. For this task, it is sufficient and necessary to consider the value of the discriminant. And when solving a quadratic equation, solving the discriminant solves the question of the nature of their solutions. Don't stop learning. I was born in 1950 and I'm still learning.😎
Sir, please upload a lecture on vectors basics and advance
a^2-(2x^2+1)a+x(x^3+1)=0
a=(2x^2+1+-sqrt(4x^4+4x^2+1-4x^4-4x))/2
a=(2x^2+1+-sqrt(4x^2-4x+1))/2
a=(2x^2+1+-(2x-1))/2
a1=x^2+x
a2=x^2-x+1
x^2+x-a1=0
x1=(-1+-sqrt(1+4a1))/2
x^2-x+(1-a2)=0
x2=(1+-sqrt(4a2-3))/2
1+4a1>=0; 4a2-3>=0
a1>=-1/4; a2>=3/4
since both must be true for all 4 roots to be real, a>=3/4
Sir can you find all real and complex solutions for this polynomial equation which has fractional powers that is
X^{4/3} - 4X^{2} + 4 = 0
Prime Newtoon's owner, how to find d^i/dx^i, when i is imaginary number?
That's an excellent question, you are referring of course to the i-th derivative operator Dⁱ (f(x)) = fⁱ (x).
From _D-operator theory_ , F(D)e^(ax) ≡ e^(ax). F(a)
so, for imaginary indices operation; Dⁱ [e^(ax)] = e^(ax). aⁱ
and for polynomial functions, Dⁱ [xⁿ] = α(n)xⁿ-ⁱ where function α(n) is complex.
there are similar formulae for trig functions.
This is the basis for Laplace operators in solving differential equations, so not so esoteric.
@@tomctutor thank you very much
You are a gifted instructor. But tell us about yourself. Where did you get your math background. You are a mystery man.
Clever solution! Thanks for sharing!
Excellent approach, one that is difficult to see.
I think jp maths made a video on this recently...
Please don't write obvious steps like -(-(2x²+1)) and then 2x²+1, that's annoying. Btw your videos are awesome.
Interesante, yo para evitarme esas cosas, simplemente aplique aspa doble especial a la ecuacion de 4to grado, me dejaron 2 expresiones de 2do grado, discriminante >=0 para ambos casos con la 'a' y finalmente intersectar
Please, can you explain what exactly the Feynman technique, with simple exercise
Thanks teacher... U relaly have a good heart
I need u as my math teacher 😭😭😭😭😭😭😭😭😭
Pls can u add numerical videos
A real beauty.
I wish I was smart enough to have earned a degree in math. The Bible verses at the end are also a nice touch.
Are you familiar with the discriminant (b^2 - 4ac)?
Don't think so
@@PrimeNewtonsIt's the part of the quadratic formula inside the square root.
@@Straight_Talk Well you live and learn!
Nice
Thanks Sir 👍
Excelente 🇧🇷 🇧🇷 🇧🇷
J pi maths did a video on this exact problem yesterday.
His equation has a -x term in place of the +x term.
@@brendanward2991And besides, he asked for only one real positive solution.
Nice
√(2x-1)^2 =|2x-1|. Modul i thing in this moment 1 little mistake 5:18
@@shj_happy10 i wont a explanation 🙂
If you use ±before root you shouldn't put || after lifting root @@just_sofi
@@shj_happy10 no i know this but more i know If under the square root there is a number squared, then the number under the modulus comes out of it because 3 square is 9 and -3 square is also 9 and √(-3)^2 =3 not -3 i don't know you understand what i explained because my English not fluently :)
@@shj_happy10noo bro she is actually correct
√x² = |x|
Because u know that's how modulus of a square root function is defined in order not to break fundamental theorem of zeroes of polynomials
So √9 is only 3, not -3
U may see it like that
√9 = √3² = |3| = 3
√9 = √(-3)² = |-3| = 3
So as per the definition we are having one answer and that's correct
@@shj_happy10 A negative number never comes out of the root. I think we are not understand each other :) thank you also have a nice day
This is a very intriguing problem. Superb solution.
ok although it seems counterproductive for a representative of math and logic to throw in that non-sequitor at the end about loving darkness due to evil deeds.
The answer doesn’t make sense. One of the conditions for a is that it has to be greater than or equal to -1/4. That should be the correct answer. If we say that the answer is a is greater than or equal to 3/4, then that eliminates -1/4 because 3/4 is greater than -1/4. But we’ve already established that a is greater than or equal to -1/4 is a condition for a. Therefore, the correct answer to the question is this: a is greater than or equal to -1/4.
If you check a = 0, you will only get two real roots (the other two are non-real), so that eliminates the condition that a is greater than or equal to -1/4. Hope this helps!
Tremendous! What a splendid strategy to attack this problem.