My approach to factor the left hand side of x⁴ + 4x² + 4x + 7 = 0 into two quadratics by transforming the left hand side into a difference of two squares would be as follows. We can complete x⁴ + 4x² into a square (x² + 2)² by adding 4, but then we also have to substract this again of course, so we get (x² + 2)² − 4 + 4x + 7 = 0 which is (x² + 2)² + 4x + 3 = 0 To turn the left hand side into a difference of two squares we start by writing this as a difference of two terms, that is, as (x² + 2)² − (−4x − 3) = 0 Now, the first term (x + 2)² at the left hand side is already a square, but the second term (−4x − 3) is not yet a square. The difference between two terms will not change if we add the same thing to both terms. So, the idea is to add something to both terms in such a way that the first term will _remain_ a square and the second term will _become_ a square as well. Taking any number k, we can add 2k(x² + 2) + k² = 2kx² + k² + 4k to both terms, since (x² + 2)² + 2k(x² + 2) + k² = (x² + 2 + k)² so the first term will remain a square regardless of the value of k. Adding 2k(x² + 2) + k² = 2kx² + k² + 4k to each of the two terms at the left hand side we have (x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0 Now, the first term at the left hand side is a square for any value of k, so we are free to choose k in such a way that the second term 2kx² − 4x + k² + 4k − 3, which is a quadratic in x, will also become a perfect square, that is, the square of a linear polynomial in x. A quadratic ax² + bx + c is a perfect square if and only if this quadratic has two identical zeros (that is, a single zero with multiplicity 2) and this is the case if and only if the discriminant b² − 4ac of this quadratic is zero. So, the condition which k must satisfy in order for the the second term 2kx² − 4x + k² + 4k − 3 to become a perfect square is (−4)² − 4·2k·(k² + 4k − 3) = 0 which gives k³ + 4k² − 3k − 2 = 0 The sum of the coefficients of this cubic in k is zero, so k = 1 is a solution of this cubic equation. With k = 1 our quartic equation (x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0 becomes (x² + 3)² − (2x² − 4x − 2) = 0 or (x² + 3)² − 2(x − 1)² = 0 or (x² + 3)² − (√2·x − √2)² = 0 which is the same result as in the video, but we only had to deal with a single equation in a single variable k to obtain this result. Using the difference of two squares identity the quartic can now be factored as (x² − √2·x + 3 + √2)(x² + √2·x + 3 − √2) = 0 The discriminants of both quadratic factors are negative, so the quartic has no real solutions. Alternatively, we could also have started by rewriting the equation as (x²)² − (−4x² − 4x − 7) = 0 and then add 2kx² + k² to both terms at the left hand side to get (x² + k)² − ((2k − 4)x² − 4x + (k² − 7)) = 0 where k must satisfy k³ − 2k² − 7k + 12 = 0 to make the second term a perfect square. This is of course the same cubic equation as the cubic equation in a in the video.
I solved it similarly (see above), according to Ferrari. The only major difference is that rather than writing the difference of two squares on one side, e.g. A^2 - B^2 = 0, and then using the 3rd binomic formula to factorize to (A + B)(A - B) = 0, I prefer to write A^2 = B^2 and then taking the square root to A = +-B, which of course leads to the same results. And I had a slightly different cubic resolvent, but that's normal. For instance, Descartes used a similar but linearly transformed cubic resolvent in comparison to Ferrari.
a^2 - 2c = 7 is wrong!? The solution can be checked with a root finder computer calculator for x^4 + 4x^2 + 4x + 7 = 0 equation. I didn't realize until looking up scientific calculators they don't have abilities to program a software computer calculation, nowadays. I still use my Muller's Method code in Basic programming language in a 1985 still functioning computer calculator to get roots (use Wolfram Alpha nowadays also by just typing in that polynomial and it's roots found).
Awesome!
Thanks!
Thank you so mcuh
Can you do a video explaining to us how to prepare to IMO ❤❤
My approach to factor the left hand side of
x⁴ + 4x² + 4x + 7 = 0
into two quadratics by transforming the left hand side into a difference of two squares would be as follows. We can complete x⁴ + 4x² into a square (x² + 2)² by adding 4, but then we also have to substract this again of course, so we get
(x² + 2)² − 4 + 4x + 7 = 0
which is
(x² + 2)² + 4x + 3 = 0
To turn the left hand side into a difference of two squares we start by writing this as a difference of two terms, that is, as
(x² + 2)² − (−4x − 3) = 0
Now, the first term (x + 2)² at the left hand side is already a square, but the second term (−4x − 3) is not yet a square. The difference between two terms will not change if we add the same thing to both terms. So, the idea is to add something to both terms in such a way that the first term will _remain_ a square and the second term will _become_ a square as well.
Taking any number k, we can add 2k(x² + 2) + k² = 2kx² + k² + 4k to both terms, since (x² + 2)² + 2k(x² + 2) + k² = (x² + 2 + k)² so the first term will remain a square regardless of the value of k. Adding 2k(x² + 2) + k² = 2kx² + k² + 4k to each of the two terms at the left hand side we have
(x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0
Now, the first term at the left hand side is a square for any value of k, so we are free to choose k in such a way that the second term 2kx² − 4x + k² + 4k − 3, which is a quadratic in x, will also become a perfect square, that is, the square of a linear polynomial in x.
A quadratic ax² + bx + c is a perfect square if and only if this quadratic has two identical zeros (that is, a single zero with multiplicity 2) and this is the case if and only if the discriminant b² − 4ac of this quadratic is zero. So, the condition which k must satisfy in order for the the second term 2kx² − 4x + k² + 4k − 3 to become a perfect square is
(−4)² − 4·2k·(k² + 4k − 3) = 0
which gives
k³ + 4k² − 3k − 2 = 0
The sum of the coefficients of this cubic in k is zero, so k = 1 is a solution of this cubic equation. With k = 1 our quartic equation
(x² + 2 + k)² − (2kx² − 4x + k² + 4k − 3) = 0
becomes
(x² + 3)² − (2x² − 4x − 2) = 0
or
(x² + 3)² − 2(x − 1)² = 0
or
(x² + 3)² − (√2·x − √2)² = 0
which is the same result as in the video, but we only had to deal with a single equation in a single variable k to obtain this result. Using the difference of two squares identity the quartic can now be factored as
(x² − √2·x + 3 + √2)(x² + √2·x + 3 − √2) = 0
The discriminants of both quadratic factors are negative, so the quartic has no real solutions.
Alternatively, we could also have started by rewriting the equation as
(x²)² − (−4x² − 4x − 7) = 0
and then add 2kx² + k² to both terms at the left hand side to get
(x² + k)² − ((2k − 4)x² − 4x + (k² − 7)) = 0
where k must satisfy
k³ − 2k² − 7k + 12 = 0
to make the second term a perfect square. This is of course the same cubic equation as the cubic equation in a in the video.
I solved it similarly (see above), according to Ferrari. The only major difference is that rather than writing the difference of two squares on one side, e.g. A^2 - B^2 = 0, and then using the 3rd binomic formula to factorize to (A + B)(A - B) = 0,
I prefer to write A^2 = B^2 and then taking the square root to A = +-B, which of course leads to the same results.
And I had a slightly different cubic resolvent, but that's normal.
For instance, Descartes used a similar but linearly transformed cubic resolvent in comparison to Ferrari.
Nice, although kinda messy!
Cute and method 2 much neater
Thank you! 😊
There is no solotion in R
I think you put the value of c is wrong please check n let me know
a^2 - 2c = 7 is wrong!? The solution can be checked with a root finder computer calculator for x^4 + 4x^2 + 4x + 7 = 0 equation. I didn't realize until looking up scientific calculators they don't have abilities to program a software computer calculation, nowadays. I still use my Muller's Method code in Basic programming language in a 1985 still functioning computer calculator to get roots (use Wolfram Alpha nowadays also by just typing in that polynomial and it's roots found).
I thought you were going to use Ferrari's method 😅
x^4 + 4x^2 + 4x + 7 = 0
x^4 = -4x^2 - 4x - 7
Add 2zx^2 + z^2:
x^4 + 2zx^2 + z^2 = (2z - 4)x^2 - 4x + (z^2 - 7)
Left side is (x^2 + z)^2
Right side is perfect square iff. D = b^2 - 4ac = 0 or b^2 = 4ac with
a = 2z - 4 = 2(z - 2)
b = -4
c = z^2 - 7
Therefore
(-4)^2= 4 * 2*(z - 2) * (z^2 - 7)
16 = 8 * (z - 2) * (z^2 - 7)
2 = (z - 2)(z^2 - 7)
2= z^3 - 2z^2 - 7z + 14
0 = z^3 - 2z^2 - 7z + 12
Cubic resolvent
z^3 + 12 = 2z^2 + 7z
z^3 + 12 = z*(2z + 7)
z = 3 is a solution, because
3^3 + 12 = 27 + 12 = 39
3*(2*3 + 7) = 3*(6 + 7) = 3*13 = 39
...
Nice
There are no real solutions for this quartic equation.