We want to factor this equation into: (x^2 + sx + d1)(x^2 - sx + d2)=0 d1 + d2 - s^2 = 0 s(d2 - d1) = -72 d1 d2 = -17 for d1=1 d2= - 17 ans s is not real.. for d1= -1 d2=17 s^2=d1+d2=16 s= -72/(d2-d1)=-72/18=-4 (x^2 -4x -1)(x^2 +4x +17)=0 (x-2)^2-5)((x+2)^2+13)=0 and the rest of the calculations are pretty simple.
Well done Descartes' method but in my opinion if we eliminate d' s then -68 = s^4 - ( 72/s)^2 By formula 4 ab = (a+b)^2 - (a-b)^2 By trial s= 4 then we can calculate d's and final answer if we take s= -4 then we obtain same result . Actually I want to say that in some cases d's are irrational but product is rational e g. equation x^4-4 x-1= 0 here d's are 1+√2 and 1-√2 and s = √2 Same case in equation x^4 +12 x+3 = 0 ( Delhi University BSC honors 1948) Here d''s are 3-√6 , 3+√6 and s = √6
@@raghvendrasingh1289Your example is one where factorization into two quadratics with integer coefficients is not possible, but where the Ferrari method used in this video still leads to a managable result. Consider the equation x^4 = qx+r with q, r integers. Add 2Px^2 + P^2 to both sides, to get (x^2 + P)^2 = 2 Px^2 + q x + P^2 + r. Choose P such that the right hand side is a quadratic expression in x. This occurs when the discriminant vanishes. Let Q=2P, this occurs when Q(Q^2+4r) = q^2, in which case the right hand side of the quartic equation can be written Q(x+q/(2Q))^2 The cubic equation for Q can always be solved, but the general algebraic expression is too complicated to be useful. Except when there is an integer root, like the example with q=4, r=1. Then the cubic equation becomes Q(Q^2+4) = 16 with an integer root Q=2. Hence P=1, and the equation becomes (x^2 + 1)^2 = 2(x+1)^2. This leads to a factorization, but not over the integers (since Q=2 is not a quadratic integer). Thus, the method used by the video author is capable of handle a much larger set of cases than the (much faster) search for factorization over the integers. But it seems like all such problems solved by him allow integer factorization.
The four solutions are as follows: 2+/-sqrt(5), -2+/-sqrt(13)i. I have three questions: 1.) at the 15: 30 mark, how did you simplify those equations? 2.) Could you make a playlist of Ferrari problems, named after Scipio Ferrari? 3.) I have noticed that there are similarities to the m^2-m^3=3/64 as well as that fourth root(97-5x)-fourth root(5x)=5. Is that also Vieta problem? Could there be playlist of Vieta problems??? I am focusing on maxing out my algebra skills using this channel.
@michaeldoerr5810: You are confusing Scipione del Ferro (1465-1526) and Lodovico Ferrari (1522-1565). See Wikipedia for their biographies and accomplishments. The solution of the quartic equation x⁴ − 72x − 17 = 0 using Ferrari's method as presented in this video is unnecessarily complicated. When we have x⁴ = 72x + 17 and note that the left hand side x⁴ = (x²)² is a perfect square we can select any number k and add 2k² + k² to both sides. Since x⁴ + 2k² + k² = (x² + k)² the left hand side will then remain a perfect square regardless of the value of the number k. So, we now have (x² + k)² = 2kx² + 72x + (k² + 17) Since the left hand side will remain a perfect square for any value of k, we are free to choose k in such a way that the right hand side will also become a perfect square. This will be the case if and only if the discrimimant of the quadratic in x at the right hand side is zero. This condition gives a cubic equation in k, but, more often than not, with quartic equations from contests, you _do not need to solve the cubic resolvent formally_ in order to proceed with the solution of the quartic equation. This is because quartic equations from contests with integer coefficients usually have nice factorizations into two quadratics with integer coefficients. Assuming a value of k exists which makes the right hand side the square of a linear polynomial in x with an _integer_ coefficient of x, this means that the coefficient 2k of x² at the right hand side will than have to be the _square of an integer_ i.e. 1, 4, 9, 16 ... So, here we only need to test values of k which make 2k the square of an integer, that is, k = ¹⁄₂, 2, ⁹⁄₂, 8 ... and check to see if this makes the constant term k² + 72 the square of an integer or the square of a rational number. With k = ¹⁄₂ we have 2k = 1 = 1² and k² + 72 = ¹⁄₄ + ²⁸⁸⁄₄ = ²⁸⁹⁄₄ = (¹⁷⁄₂)² which is the square of a rational number. However, twice the product of 1 and ¹⁷⁄₂ is 17, not 72 as it should in order for the right hand side to be a perfect square. But then, you will quickly find that k = 8 _does_ fit the bill, because this makes 2k = 16 = 4² as well as k² + 17 = 64 + 17 = 81 = 9² while the coefficient 72 of the linear term at the right hand side is indeed twice the product of 4 and 9. So, we select k = 8 and our equation then becomes (x² + 8)² = 16x² + 72x + 81 which can be written as (x² + 8)² = (4x + 9)² and we have accomplished our goal of rewriting the original equation in such a way that both sides are a perfect square. We can now bring the perfect square from the right hand side over to the left hand side to get (x² + 8)² − (4x + 9)² = 0 At the left hand side we now have a difference of two squares where we can apply the difference of two squares identity a² − b² = (a − b)(a + b) to get (x² + 8 − 4x − 9)(x² + 8 + 4x + 9) = 0 or (x² − 4x − 1)(x² + 4x + 17) = 0 Applying the zero product property which says that a product is zero if and only if at least one of its factors is itself zero this gives x² − 4x − 1 = 0 ⋁ x² + 4x + 17 = 0 and now we only need to solve these two quadratic equations to find all four solutions of our original quartic equation.
This is not a good method because it relies on an 'element of luck.' If you had written it as L.H.S.=(x^2+ax-17)(x^2-ax+1)=0, you might not have found the solution. I described the correct method belowe a different, similar problem. The correct method is L.H.S.=(x^2+ax+d1)(x^2-ax+d2)=0, wher d1 *d2= Constant term. This method works when the polynomial L.H.S can be factored into quadratic factors with integer coefficients, and it is easy when the coefficient of x^3 in the left-hand side (LHS) polynomial is zero.
In almost all these videos there is a step missing. X^4 is a complete square by itself. So what is going on is if m = x^2 is substituted into (m+n)^2 then (m+n)^2 expands out to the LHS. This is no longer simply X^4. But the RHS is what X^4 is equal to (ie 72x + 17), so all terms on the LHS additional to X^4 need to be added to 72x + 17 to make the equations balance.
@@superacademy247 @superacademy247 It's a bit hard to describe, but I will try. I learned this method from another math RUclipsr, blackpenredpen. His method starts by assuming the quartic will nicely factor into 2 quadratics. Assuming this is the case, the x^2 term of each quadratic is obviously going to be just x^2 with no coefficient. In order to get the constant terms of each quadratic, that's also pretty simple - 17 is prime, so the only possibilities to produce a product of -17 are 1 and -17 or -1 and 17. For composite numbers, this is a bit harder. But you basically want to guess and check each of these possibilities, seeing if they satisfy the x^1 terms. The next step is to split the quartic into 5 columns, one for each exponent of x. If you need to add in terms with coefficients of 0 to do this, which is the case in this problem, you should. Our quartic now looks like this: x^4 | +0x^3 | +0x^2 | -72x | -17 Next, you should add 2 empty rows to the bottom of each column. These will be important later. For the first column, you should fill in the spaces with your x^2 terms for each quadratic. For the third column, you should fill the spaces with ax and bx. a and b are variables we will solve for later. For the last column, fill the spaces with the two constant factors you are guessing. It doesn't really matter which factor goes in which row, but you might want to stay consistent, especially if you are dealing with a composite constant term. The grid now looks like this (for this problem): x^4 | +0x^3 | +0x^2 | -72x | -17 x^2 | | ax | | -1 x^2 | | bx | | 17 In the remaining spaces, you should put the products of the terms that are knight's moves away from each other (within the bottom two rows). Once again, it doesn't really matter what row you put each product in, but make sure the products are between their respective terms, and that you do this in a symmetrical way. The rows will now look like this (I will remove the top row, as it is annoying to type out) x^2 | bx^3 | ax | -bx | -1 x^2 | ax^3 | bx | 17ax | 17 Then, for the second and fourth columns, add the products and set them equal to their respective terms in the equation above. Set these two equations off to the side, making sure to cancel out the x^3 and x^1 terms. You now have a system of equations to solve for a and b. The system of equations for this problem is as follows: -17a + b = -72 (column 4) a + b = 0 (column 2) Solve this system of equations for a and b. Plug their values into the third column. To check if they are the correct coefficients of the x^1 terms in the quadratics, multiply them together, and add the product of the first x^2 term of the quadratics multiplied by the second constant term of the quadratics, and the product of the reverse of that. To clarify, this is what you should be adding: abx^2 + 17x^2 + (-x^2) If their sum is equal to the coefficient of the x^2 term in the original equation, you have succeeded. Your quadratic factors are (x^2 + ax - 1) and (x^2 + bx + 17). If you did the method correctly, your a and b values should be -4 and 4. If their sum is not equal, it means your initial guess of the constant factors was wrong, and you will have to make a new guess, applying the same process to your new guess. Hopefully my explanation is sufficient.
If x^^4 + ax² +bx + c = 0 (you can have no term with x^^3 with an appropriate translation) you can introduce y as a number and write : (x² + y)² - [(2y-a)x² - bx + y² - c] = 0 If the second terme is a square we have delta = b² - 4(2y-a)(y²-c) = 0 and you have to solve a 3 degree equation to choose y In this case a = 0 b = -72 c = -17 so 5184 - 8(y^^3 + 17y) = 0 or 648 - y^^3 - 17y = 0 8 is a solution so you have : (x² + 8)² - (16x² + 72x + 81) = 0 or (x²+ 8)² - (4x + 9)² =0 or (x² + 4x + 17)(x² -4x -1) = 0
We want to factor this equation into:
(x^2 + sx + d1)(x^2 - sx + d2)=0
d1 + d2 - s^2 = 0
s(d2 - d1) = -72
d1 d2 = -17
for d1=1 d2= - 17 ans s is not real..
for d1= -1 d2=17
s^2=d1+d2=16
s= -72/(d2-d1)=-72/18=-4
(x^2 -4x -1)(x^2 +4x +17)=0
(x-2)^2-5)((x+2)^2+13)=0
and the rest of the calculations are pretty simple.
Well done
Descartes' method but in my opinion if we eliminate d' s then
-68 = s^4 - ( 72/s)^2
By formula 4 ab = (a+b)^2 - (a-b)^2
By trial s= 4 then we can calculate d's and final answer if we take s= -4 then we obtain same result .
Actually I want to say that in some cases d's are irrational but product is rational
e g. equation x^4-4 x-1= 0
here d's are 1+√2 and 1-√2 and s = √2
Same case in equation x^4 +12 x+3 = 0 ( Delhi University BSC honors 1948)
Here d''s are 3-√6 , 3+√6 and s = √6
@@raghvendrasingh1289Your example is one where factorization into two quadratics with integer coefficients is not possible, but where the Ferrari method used in this video still leads to a managable result. Consider the equation
x^4 = qx+r
with q, r integers. Add 2Px^2 + P^2 to both sides, to get
(x^2 + P)^2 = 2 Px^2 + q x + P^2 + r.
Choose P such that the right hand side is a quadratic expression in x. This occurs when the discriminant vanishes. Let Q=2P, this occurs when
Q(Q^2+4r) = q^2,
in which case the right hand side of the quartic equation can be written
Q(x+q/(2Q))^2
The cubic equation for Q can always be solved, but the general algebraic expression is too complicated to be useful. Except when there is an integer root, like the example with q=4, r=1. Then the cubic equation becomes
Q(Q^2+4) = 16
with an integer root Q=2. Hence P=1, and the equation becomes
(x^2 + 1)^2 = 2(x+1)^2.
This leads to a factorization, but not over the integers (since Q=2 is not a quadratic integer).
Thus, the method used by the video author is capable of handle a much larger set of cases than the (much faster) search for factorization over the integers. But it seems like all such problems solved by him allow integer factorization.
The four solutions are as follows: 2+/-sqrt(5), -2+/-sqrt(13)i. I have three questions: 1.) at the 15: 30 mark, how did you simplify those equations? 2.) Could you make a playlist of Ferrari problems, named after Scipio Ferrari? 3.) I have noticed that there are similarities to the m^2-m^3=3/64 as well as that fourth root(97-5x)-fourth root(5x)=5. Is that also Vieta problem? Could there be playlist of Vieta problems??? I am focusing on maxing out my algebra skills using this channel.
I'll make that playlist for easy reference 😎😍. Thanks 👍💯.
@michaeldoerr5810: You are confusing Scipione del Ferro (1465-1526) and Lodovico Ferrari (1522-1565). See Wikipedia for their biographies and accomplishments.
The solution of the quartic equation x⁴ − 72x − 17 = 0 using Ferrari's method as presented in this video is unnecessarily complicated. When we have
x⁴ = 72x + 17
and note that the left hand side x⁴ = (x²)² is a perfect square we can select any number k and add 2k² + k² to both sides. Since x⁴ + 2k² + k² = (x² + k)² the left hand side will then remain a perfect square regardless of the value of the number k. So, we now have
(x² + k)² = 2kx² + 72x + (k² + 17)
Since the left hand side will remain a perfect square for any value of k, we are free to choose k in such a way that the right hand side will also become a perfect square. This will be the case if and only if the discrimimant of the quadratic in x at the right hand side is zero. This condition gives a cubic equation in k, but, more often than not, with quartic equations from contests, you _do not need to solve the cubic resolvent formally_ in order to proceed with the solution of the quartic equation. This is because quartic equations from contests with integer coefficients usually have nice factorizations into two quadratics with integer coefficients.
Assuming a value of k exists which makes the right hand side the square of a linear polynomial in x with an _integer_ coefficient of x, this means that the coefficient 2k of x² at the right hand side will than have to be the _square of an integer_ i.e. 1, 4, 9, 16 ...
So, here we only need to test values of k which make 2k the square of an integer, that is, k = ¹⁄₂, 2, ⁹⁄₂, 8 ... and check to see if this makes the constant term k² + 72 the square of an integer or the square of a rational number. With k = ¹⁄₂ we have 2k = 1 = 1² and k² + 72 = ¹⁄₄ + ²⁸⁸⁄₄ = ²⁸⁹⁄₄ = (¹⁷⁄₂)² which is the square of a rational number. However, twice the product of 1 and ¹⁷⁄₂ is 17, not 72 as it should in order for the right hand side to be a perfect square. But then, you will quickly find that k = 8 _does_ fit the bill, because this makes 2k = 16 = 4² as well as k² + 17 = 64 + 17 = 81 = 9² while the coefficient 72 of the linear term at the right hand side is indeed twice the product of 4 and 9. So, we select k = 8 and our equation then becomes
(x² + 8)² = 16x² + 72x + 81
which can be written as
(x² + 8)² = (4x + 9)²
and we have accomplished our goal of rewriting the original equation in such a way that both sides are a perfect square. We can now bring the perfect square from the right hand side over to the left hand side to get
(x² + 8)² − (4x + 9)² = 0
At the left hand side we now have a difference of two squares where we can apply the difference of two squares identity
a² − b² = (a − b)(a + b)
to get
(x² + 8 − 4x − 9)(x² + 8 + 4x + 9) = 0
or
(x² − 4x − 1)(x² + 4x + 17) = 0
Applying the zero product property which says that a product is zero if and only if at least one of its factors is itself zero this gives
x² − 4x − 1 = 0 ⋁ x² + 4x + 17 = 0
and now we only need to solve these two quadratic equations to find all four solutions of our original quartic equation.
At 14.22 , you should take sq. root for both sides
Graphing on hp-prime shows the two real roots (as I have it set up) and wolframalpha also gives the two complex roots.
That is it!
I know another method which is too easier than this.
Let L.H.S.=(x^2+ax+17)(x^2-ax-1)=0......etc
This is not a good method because it relies on an 'element of luck.' If you had written it as
L.H.S.=(x^2+ax-17)(x^2-ax+1)=0, you might not have found the solution. I described the correct method belowe a different, similar problem.
The correct method is L.H.S.=(x^2+ax+d1)(x^2-ax+d2)=0, wher d1 *d2= Constant term.
This method works when the polynomial L.H.S can be factored into quadratic factors with integer coefficients, and it is easy when the coefficient of x^3 in the left-hand side (LHS) polynomial is zero.
In almost all these videos there is a step missing. X^4 is a complete square by itself. So what is going on is if m = x^2 is substituted into (m+n)^2 then (m+n)^2 expands out to the LHS. This is no longer simply X^4. But the RHS is what X^4 is equal to (ie 72x + 17), so all terms on the LHS additional to X^4 need to be added to 72x + 17 to make the equations balance.
No step missing. I think there's a concept you're missing especially completing the square method
This method seems far more complicated than the method I used. Not to be rude, but this method is atrocious, and I will likely never use it.
Please share your method
@@superacademy247 @superacademy247 It's a bit hard to describe, but I will try.
I learned this method from another math RUclipsr, blackpenredpen.
His method starts by assuming the quartic will nicely factor into 2 quadratics.
Assuming this is the case, the x^2 term of each quadratic is obviously going to be just x^2 with no coefficient.
In order to get the constant terms of each quadratic, that's also pretty simple - 17 is prime, so the only possibilities to produce a product of -17 are 1 and -17 or -1 and 17. For composite numbers, this is a bit harder. But you basically want to guess and check each of these possibilities, seeing if they satisfy the x^1 terms.
The next step is to split the quartic into 5 columns, one for each exponent of x. If you need to add in terms with coefficients of 0 to do this, which is the case in this problem, you should.
Our quartic now looks like this:
x^4 | +0x^3 | +0x^2 | -72x | -17
Next, you should add 2 empty rows to the bottom of each column. These will be important later.
For the first column, you should fill in the spaces with your x^2 terms for each quadratic. For the third column, you should fill the spaces with ax and bx. a and b are variables we will solve for later. For the last column, fill the spaces with the two constant factors you are guessing. It doesn't really matter which factor goes in which row, but you might want to stay consistent, especially if you are dealing with a composite constant term.
The grid now looks like this (for this problem):
x^4 | +0x^3 | +0x^2 | -72x | -17
x^2 | | ax | | -1
x^2 | | bx | | 17
In the remaining spaces, you should put the products of the terms that are knight's moves away from each other (within the bottom two rows). Once again, it doesn't really matter what row you put each product in, but make sure the products are between their respective terms, and that you do this in a symmetrical way. The rows will now look like this (I will remove the top row, as it is annoying to type out)
x^2 | bx^3 | ax | -bx | -1
x^2 | ax^3 | bx | 17ax | 17
Then, for the second and fourth columns, add the products and set them equal to their respective terms in the equation above. Set these two equations off to the side, making sure to cancel out the x^3 and x^1 terms. You now have a system of equations to solve for a and b. The system of equations for this problem is as follows:
-17a + b = -72 (column 4)
a + b = 0 (column 2)
Solve this system of equations for a and b. Plug their values into the third column. To check if they are the correct coefficients of the x^1 terms in the quadratics, multiply them together, and add the product of the first x^2 term of the quadratics multiplied by the second constant term of the quadratics, and the product of the reverse of that. To clarify, this is what you should be adding:
abx^2 + 17x^2 + (-x^2)
If their sum is equal to the coefficient of the x^2 term in the original equation, you have succeeded. Your quadratic factors are (x^2 + ax - 1) and (x^2 + bx + 17). If you did the method correctly, your a and b values should be -4 and 4.
If their sum is not equal, it means your initial guess of the constant factors was wrong, and you will have to make a new guess, applying the same process to your new guess.
Hopefully my explanation is sufficient.
If x^^4 + ax² +bx + c = 0 (you can have no term with x^^3 with an appropriate translation) you can introduce y as a number and write :
(x² + y)² - [(2y-a)x² - bx + y² - c] = 0
If the second terme is a square we have delta = b² - 4(2y-a)(y²-c) = 0 and you have to solve a 3 degree equation to choose y
In this case a = 0 b = -72 c = -17 so 5184 - 8(y^^3 + 17y) = 0 or 648 - y^^3 - 17y = 0 8 is a solution so you have :
(x² + 8)² - (16x² + 72x + 81) = 0 or (x²+ 8)² - (4x + 9)² =0 or (x² + 4x + 17)(x² -4x -1) = 0
Amazing method 👌🥰
x^4 - 72x - 17 = (x^2 + a)^2 - b(x + c)^2 = x^4 + (2a - b)x^2 - 2bcx + (a^2 - bc^2)
=> 2a - b = 0 , -2bc = -72 , a^2 - bc^2 = -17
=> a^2 - bc^2 = (b/2)^2 - b*(36/b)^2 = -17
=> b^3 + 68b - 5184 = (b - 16)(b^2 + 16b + 324) = 0 => b =16, others not real.
=> b = 16, a = 8, c = 9/4
=> x^4 - 72x - 17 = (x^2 + 8)^2 - 16(x + 9/4)^2 = (x^2 + 8)^2 - (4x + 9)^2
=> (x^2 + 4x + 17)(x^2 - 4x -1) = 0 => x = ...