You can actually "cheese" through the problem once you noticed that 182 = 13² + 13. You can then re-write this equation " *in 13* " (yep, sounds silly, but treat "13" as a variable) - it will be "quadratic in 13". It ends up like this: 13² + (2x²+1)13 + (x⁴-x) = 0, leading to 13 = (-(2x²+1)±√4x²+4x+1)/2 , so 13 = (-(2x²+1)±(2x+1))/2 and this yields just two quadratics in *x*
The second method is really the quartic formula in disguise. Assume the equation is x^4+px^2+qx+r=0, and set the polynomial equal to (x^2+ax+b)(x^2-ax+c) as you say above. This gives b+c-a^2=p, a(c-b)=q, and bc=r. This means that (c+b)^2-(c-b)^2 = (c+b+c-b)(c+b-c+b) = 2c(2b) = 4bc = 4r. But c+b= p+a^2, and c-b = q/a. This means that (p+a^2)^2-(q/a)^2 = 4r. Multiply by a^2 and get a^2(p+a^2)^2-q^2 = 4a^2r. Simplifying this equation and letting y = a^2 yields y^3+2py^2+(p^2-4r)y-q^2=0, which is the cubic resolvent, the central part of the quartic formula.
The first approach is neat. You can streamline the argument a bit if you realise that if functions f and f inverse are equal, it can only be at a point where they both equal x. So you can just set x^2 + 13 = x and solve right away. Then the other two roots can be got by dividing the quartic by that quadratic. (BTW, I think there's a little consistency problem in the first approach. You argued that x - 13 > 0 allows you to take a positive square root, but then get complex values for x.)
This is incorrect regarding a function and its inverse. Consider the function f(x)=√(7-3x), it intersects its inverse function at (1,2) and (2,1) in addition to where √(7-3x)=x.
If you want to factor a quartic into two quadratics there are smarter ways than what you do in your second method. The idea is to transform the quartic into a difference of two squares where one square is the square of a quadratic polynomial whereas the other square is the square of a linear polynomial. Then, when we use the difference of two squares identity we get a product of two quadratics and we have achieved what we wanted. The equation is x⁴ + 26x² − x + 182 = 0 As you already noted we have x⁴ + 26x² + 169 = (x² + 13)² so we can write the equation as (x² + 13)² − (x − 13) = 0 Now, the first term is already a square of a quadratic polynomial, but (x − 13) is not a square of a linear polynomial. But what we can do here is introduce a parameter t and add 2(x² + 13)t + t² because (x² + 13)² + 2(x² + 13)t + t² = (x² + 13 + t)² so the first term will then remain a square of a quadratic polynomial regardless of the value of t. But of course if we add 2(x² + 13)t + t² = 2tx² + 26t + t² we also need to subtract this again, so the equation then becomes (x² + 13 + t)² − (2tx² + x + t² + 26t − 13) = 0 Now, for any value of t unequal to zero 2tx² + x + t² + 26t − 13 is a quadratic polynomial in x, and this will be a perfect square, i.e. the square of a linear polynomial in x, if the discriminant of this quadratic is zero, that is, if t satisfies 1² − 4·2t·(t² + 26t − 13) = 0 This is a cubic equation in t, but rather than first expanding the left hand side it is a good idea to start looking for rational solutions right away. It is clear that the left hand side is equal to 1 for t = 0 and negative for t = 1 so there must be a root on the interval [0, 1]. We may also note that 26t − 13 = 0 for t = ½ and for t = ½ we therefore get 1 − 4·2·½·¼ = 0, so t = ½ is indeed a solution of this equation in t. With t = ½ our quartic equation becomes (x² + 13 + ½)² − (x² + x + ¼) = 0 and indeed x² + x + ¼ = (x + ½)² is a perfect square, so we have (x² + 13½)² − (x + ½)² = 0 Applying the difference of two squares identity this can be written as (x² − x + 13)(x² + x + 14) = 0 and we have factored the quartic into two quadratics.
X^4+26x^2-x+182=0 For real values of x the above equation is greater than zero. x is complex number. x^4+26x^2+182=x x^4+27x^2+182=x^2+x (x^2+13)(x^2+14)=x(x+1) x^2+13=x x^2-x+13=0 x={1±(√1-52)}/2
x⁴+26x²-x+182=0 No real roots: Note that x = x⁴+26x²+182≥182. Clearly for x ≥ 1, x⁴+26x²-x+182 ≥ x⁴-x ≥ 0. Therefore if this factorizes, we're looking for (x²+ax+c)(x²-ax+d)=0 (The a/-a are forced because the coefficient of x³ is 0.) So: -a²+c+d=26 a(d-c)=-1 cd=182 If this factorizes "nicely", then a,c,d are integers, so a=±1, so c+d=27, c-d=±1, {c,d}={13,14}, which works. => ... = (x²+x+14)(x²-x+13) => x = (-1±√55i)/2 or (1±√51i)/2
@12:30 You could have just jumped to this equation immediately by noticing that the right hand side of the original equation is always greater than zero and so there are four complex roots so it factors as two quadratic equations whose discriminates are complex, which when multiplied out, has no x^3 term.
An explanation to second method : if u input c+b-a²=26 = eqn1 , a(c-b)=-1=eqn2 and bc=182=eqn3 in wolfram you'll get abc = (-1,13,14)(1,14,13) and 4 other complex sets which we dont care , so try eliminating b first then c , cause a is easiest to find , by substituting b=182/c in eqn2 you'll see (a)c² + c +(-182a) = 0 , which is a quadratic eq where c = (-1±√(1+728a²))/(2a) , putting that in eqn1 , you will quickly notice √(1+728a²) must be √729 cuz its perfect square so a must be 1 and later you can find b and c
x^4 +26 x^2 - x + 182 = 0 can be rewritten as (x^2 +13)^2 - x +13 = 0 or x^4 + 2*13*x^2 - x + 13*14 = 0 Coefficient of x^3 being zero, this can be rewritten as (x^2 +bx + 13)( x^2 - bx + 13+1) = 0 comparing coefficients of x one gets 13*b - (13+1)b = -1 i.e. b =1 comparing coefficients of x^2 one gets 13+1 +13 -b*b = 13*2 i.e b*b =1 (which is in compliance of b=1) Hereby, given equation can be rewritten as (x^2 +x + 13)( x^2 - x + 13+1) = 0
the eqution is written as X^2 (X^2 + 26) = x - 182. The LHS is a U-shaped curve touching at (0,0), the RHS is a straight line cutting the y-axis at --182 with slope = 1, they do not intersect, we therefore know before solving, the equattion has two pairs of complex numbers
(x^2+13)^2-x+13=0 Let 13=a and substitute. It will genarate a quadratic equation in a. Solve with quadratic formula and re substute 13 for a to generate two quadratic equations of x. Solve them for the answer. Third method.
Can you explain 2 things for me? First, when I put x^2+13=(x-13)^(1/2) in my calculator it says there are no solutions for the equation. How is that so, even though it's derived from the initial equation? Is it because the 'x' on the right is caught under the square root symbol? Second, after you got to (x-y)(1+x+y)=0, how did you come to determine that x-y=0? I heard something earlier about the inverse functions being symmetric where x=y or something like that. Can you elaborate? Thank you. This is a great video! Edit: I understand the second part lol.
Your calculator is right, there are no solutions because the two equations do not intersect at any point. Think about it, you have a parabola starting at a height of 13. And a square root function moved to the right 13 units. Which means you can only start at x=13 but if you start at x=13 the parabola starts at a height of 182 and grows way faster then the square root function and therefore will never intersect. By the way when x=13 the square root function starts at 0.
Write it as x^4+27x^2-x^2-x+182=0 x^2(x^2+27)-(x^2+x-182)=0 x^2(x^2+27)=(x^2+x-182) x^2(x^2+27)=(x-13)(x+14) Now, we know that x^2+27>x^2 So, on comparing, we get, x^2=x-13 and x^2+27=x+14 We get, x^2-x+13=0 on both equations. That means we are right so far. Solving for x using quadratic formula, we get, x = {1+/-sqrt(1-52)}/2 x = {1 +/- sqrt(-51)}/2 x = (1+sqrt51*i)/2 and (1-sqrt51*i)/2 Hence, our question is solved.
in 3:23 you have (x²+13)^2 = x-13 being the right hand side non negative for x greater than 13, we expect to get real solutions from then on. But the function on the left hand side has derivative grater than the function on the right hand side, and, since its value at x=13 is greater than 0 (the value of the line at 13), there can be no solutions.
well, here is a very simple method : add and subtract same terms (x^2 + 13)^2 - x + 13 = (x^2 + 13)^2 - x^2 + x^2 - x + 13 = (x^2 + x + 13)(x^2 - x + 13) + (x^2 - x + 13) = (x^2 - x + 13)(x^2 + x + 13 + 1) = (x^2 - x + 13)(x^2 + x + 14) why did not I notice...
There's no real solution. x = 182 + 26x² + x⁴. If x < 182, the right is greater. If x >= 182, the right swamps x. I couldn't go any further, except that I did notice that 182 = 169 + 13 = 13² + 13, as the top answer begins.
@@SyberMath I got x equals - square root of n26ntimes 3ndivided byn2..using quadratic formula tosolve for x squared..this is valid too..do you see what I mean?
@@SyberMath i am stuck at a problem can you help me out.Its, If the area of an arbitary triangle is 'A' then find the maximum possible value of its perimeter.
Problem is that the roots of this equation are expressed with complex radicals so in real numbers they can be expressed with trigonometric functions only The conclusion is that this quartic will not help with your problem
You can actually "cheese" through the problem once you noticed that 182 = 13² + 13. You can then re-write this equation " *in 13* " (yep, sounds silly, but treat "13" as a variable) - it will be "quadratic in 13". It ends up like this: 13² + (2x²+1)13 + (x⁴-x) = 0, leading to 13 = (-(2x²+1)±√4x²+4x+1)/2 , so 13 = (-(2x²+1)±(2x+1))/2 and this yields just two quadratics in *x*
That is so cool! This is a great idea used in some algebra problems!!!
Incredible❗
Never thought about that.
Woow 👍👌👌👌
COOL!!
The second method is really the quartic formula in disguise. Assume the equation is x^4+px^2+qx+r=0, and set the polynomial equal to (x^2+ax+b)(x^2-ax+c) as you say above. This gives b+c-a^2=p, a(c-b)=q, and bc=r. This means that (c+b)^2-(c-b)^2 = (c+b+c-b)(c+b-c+b) = 2c(2b) = 4bc = 4r. But c+b= p+a^2, and c-b = q/a. This means that (p+a^2)^2-(q/a)^2 = 4r. Multiply by a^2 and get a^2(p+a^2)^2-q^2 = 4a^2r. Simplifying this equation and letting y = a^2 yields y^3+2py^2+(p^2-4r)y-q^2=0, which is the cubic resolvent, the central part of the quartic formula.
Very nice!
If you can obtain a rational root for the resolvent cubic equation, you can reduce the amount of work needed to solve the quartic.
The first approach is neat. You can streamline the argument a bit if you realise that if functions f and f inverse are equal, it can only be at a point where they both equal x. So you can just set x^2 + 13 = x and solve right away. Then the other two roots can be got by dividing the quartic by that quadratic. (BTW, I think there's a little consistency problem in the first approach. You argued that x - 13 > 0 allows you to take a positive square root, but then get complex values for x.)
Thank you!
This is incorrect regarding a function and its inverse. Consider the function f(x)=√(7-3x), it intersects its inverse function at (1,2) and (2,1) in addition to where √(7-3x)=x.
If you want to factor a quartic into two quadratics there are smarter ways than what you do in your second method. The idea is to transform the quartic into a difference of two squares where one square is the square of a quadratic polynomial whereas the other square is the square of a linear polynomial. Then, when we use the difference of two squares identity we get a product of two quadratics and we have achieved what we wanted.
The equation is
x⁴ + 26x² − x + 182 = 0
As you already noted we have x⁴ + 26x² + 169 = (x² + 13)² so we can write the equation as
(x² + 13)² − (x − 13) = 0
Now, the first term is already a square of a quadratic polynomial, but (x − 13) is not a square of a linear polynomial. But what we can do here is introduce a parameter t and add 2(x² + 13)t + t² because (x² + 13)² + 2(x² + 13)t + t² = (x² + 13 + t)² so the first term will then remain a square of a quadratic polynomial regardless of the value of t. But of course if we add 2(x² + 13)t + t² = 2tx² + 26t + t² we also need to subtract this again, so the equation then becomes
(x² + 13 + t)² − (2tx² + x + t² + 26t − 13) = 0
Now, for any value of t unequal to zero 2tx² + x + t² + 26t − 13 is a quadratic polynomial in x, and this will be a perfect square, i.e. the square of a linear polynomial in x, if the discriminant of this quadratic is zero, that is, if t satisfies
1² − 4·2t·(t² + 26t − 13) = 0
This is a cubic equation in t, but rather than first expanding the left hand side it is a good idea to start looking for rational solutions right away. It is clear that the left hand side is equal to 1 for t = 0 and negative for t = 1 so there must be a root on the interval [0, 1]. We may also note that 26t − 13 = 0 for t = ½ and for t = ½ we therefore get 1 − 4·2·½·¼ = 0, so t = ½ is indeed a solution of this equation in t. With t = ½ our quartic equation becomes
(x² + 13 + ½)² − (x² + x + ¼) = 0
and indeed x² + x + ¼ = (x + ½)² is a perfect square, so we have
(x² + 13½)² − (x + ½)² = 0
Applying the difference of two squares identity this can be written as
(x² − x + 13)(x² + x + 14) = 0
and we have factored the quartic into two quadratics.
X^4+26x^2-x+182=0
For real values of x the above equation is greater than zero.
x is complex number.
x^4+26x^2+182=x
x^4+27x^2+182=x^2+x
(x^2+13)(x^2+14)=x(x+1)
x^2+13=x
x^2-x+13=0
x={1±(√1-52)}/2
I enjoy youŕ classes. Congratulations.
Thank you! 😃
@Иван Пожидаев when f(x)=x
Why did you put an accent character over the letter r?
x⁴+26x²-x+182=0
No real roots: Note that x = x⁴+26x²+182≥182. Clearly for x ≥ 1, x⁴+26x²-x+182 ≥ x⁴-x ≥ 0.
Therefore if this factorizes, we're looking for
(x²+ax+c)(x²-ax+d)=0
(The a/-a are forced because the coefficient of x³ is 0.) So:
-a²+c+d=26
a(d-c)=-1
cd=182
If this factorizes "nicely", then a,c,d are integers, so a=±1, so c+d=27, c-d=±1, {c,d}={13,14}, which works.
=> ... = (x²+x+14)(x²-x+13)
=> x = (-1±√55i)/2 or (1±√51i)/2
I think it is "good" to mention that 182=2x91 so we can real quick assign 1, -1, 2,- 2, 91, -91 , 182 , -182 and see that there are no rational roots.
Good!
@12:30 You could have just jumped to this equation immediately by noticing that the right hand side of the original equation is always greater than zero and so there are four complex roots so it factors as two quadratic equations whose discriminates are complex, which when multiplied out, has no x^3 term.
Nice!
An explanation to second method : if u input c+b-a²=26 = eqn1 , a(c-b)=-1=eqn2 and bc=182=eqn3 in wolfram you'll get abc = (-1,13,14)(1,14,13) and 4 other complex sets which we dont care , so try eliminating b first then c , cause a is easiest to find , by substituting b=182/c in eqn2 you'll see (a)c² + c +(-182a) = 0 , which is a quadratic eq where c = (-1±√(1+728a²))/(2a) , putting that in eqn1 , you will quickly notice √(1+728a²) must be √729 cuz its perfect square so a must be 1 and later you can find b and c
x^4 +26 x^2 - x + 182 = 0
can be rewritten as
(x^2 +13)^2 - x +13 = 0
or x^4 + 2*13*x^2 - x + 13*14 = 0
Coefficient of x^3 being zero, this can be rewritten as
(x^2 +bx + 13)( x^2 - bx + 13+1) = 0
comparing coefficients of x one gets
13*b - (13+1)b = -1 i.e. b =1
comparing coefficients of x^2 one gets
13+1 +13 -b*b = 13*2 i.e b*b =1
(which is in compliance of b=1)
Hereby, given equation can be rewritten as
(x^2 +x + 13)( x^2 - x + 13+1) = 0
the eqution is written as X^2 (X^2 + 26) = x - 182. The LHS is a U-shaped curve touching at (0,0), the RHS is a straight line cutting the y-axis at --182 with slope = 1, they do not intersect,
we therefore know before solving, the equattion has two pairs of complex numbers
(x^2 +13)( x^2 + 14) - x^2 -x
= (x^2 +ax + 13)( x^2 + bx + 14), say
Then ab = -1 , a + b = 0, 14 a + 13 b = -1
Hereby a = -1 , b = 1
Therefore, given equation becomes
= (x^2 -x + 13)( x^2 + x + 14) = 0.
or x = ( 1 + √(51)) /2 , ( 1 -√(51)) /2 ,
= ( -1 + √(55)) /2 , (- 1 -√(55)) /2 ,
(x^2+13)^2-x+13=0
Let 13=a and substitute. It will genarate a quadratic equation in a. Solve with quadratic formula and re substute 13 for a to generate two quadratic equations of x. Solve them for the answer. Third method.
You should finish the second method off too bro, you know we ain't gonna do it "at home" 🤣 btw awesome 👏🏿👏🏿👏🏿
Sorry, bro! 😁
Thanks!
The decomposition of bi-quadratic into product of two quadratics is called Descartes method.
I solved...
Can you explain 2 things for me?
First, when I put x^2+13=(x-13)^(1/2) in my calculator it says there are no solutions for the equation. How is that so, even though it's derived from the initial equation? Is it because the 'x' on the right is caught under the square root symbol?
Second, after you got to (x-y)(1+x+y)=0, how did you come to determine that x-y=0? I heard something earlier about the inverse functions being symmetric where x=y or something like that. Can you elaborate?
Thank you. This is a great video!
Edit: I understand the second part lol.
A radical equation may not have the same roots as its polynomial counterpart for the reason you mentioned, the square root!
Thank YOU!!!
Your calculator is right, there are no solutions because the two equations do not intersect at any point. Think about it, you have a parabola starting at a height of 13. And a square root function moved to the right 13 units. Which means you can only start at x=13 but if you start at x=13 the parabola starts at a height of 182 and grows way faster then the square root function and therefore will never intersect. By the way when x=13 the square root function starts at 0.
Sir, I have to ask you a single question -- " Where do you bring such amezing trick and equations "
Please answer me
🙏🙏🙂😊☺️
11:05 How do you check if polynomials have real solutions? Are you referring to the descartes rule of signs+ some other tricks?
Write it as
x^4+27x^2-x^2-x+182=0
x^2(x^2+27)-(x^2+x-182)=0
x^2(x^2+27)=(x^2+x-182)
x^2(x^2+27)=(x-13)(x+14)
Now, we know that x^2+27>x^2
So, on comparing, we get,
x^2=x-13 and x^2+27=x+14
We get,
x^2-x+13=0 on both equations.
That means we are right so far.
Solving for x using quadratic formula, we get,
x = {1+/-sqrt(1-52)}/2
x = {1 +/- sqrt(-51)}/2
x = (1+sqrt51*i)/2 and (1-sqrt51*i)/2
Hence, our question is solved.
in 3:23 you have (x²+13)^2 = x-13 being the right hand side non negative for x greater than 13, we expect to get real solutions from then on. But the function on the left hand side has derivative grater than the function on the right hand side, and, since its value at x=13 is greater than 0 (the value of the line at 13), there can be no solutions.
Nice
well, here is a very simple method : add and subtract same terms
(x^2 + 13)^2 - x + 13
= (x^2 + 13)^2 - x^2 + x^2 - x + 13
= (x^2 + x + 13)(x^2 - x + 13) + (x^2 - x + 13)
= (x^2 - x + 13)(x^2 + x + 13 + 1)
= (x^2 - x + 13)(x^2 + x + 14)
why did not I notice...
Wow bro
You can use greaffe's root square method
How?
@@SyberMath this method is numerical and iterative. So it can't be show here
Brilliant!!!
Could you please solve the quartic equation 2x^4+12x^3+2x^2-72x-63=0 the equation you derived in solving a geometrical problem.
I will take a look and get back to you on that
Here is the solution:
www.wolframalpha.com/input/?i2d=true&i=2Power%5Bx%2C4%5D%2B12Power%5Bx%2C3%5D%2B2Power%5Bx%2C2%5D-72x-63%3D0
In method 2 at 12 minutes, I notice 182=13x14 , Let b=13, c=14 . Is it any better?
That would be good
@@SyberMath I get in touch your video recently and fall in love with it
Both methods are interesting. Good presentation.
Glad you think so!
There's no real solution. x = 182 + 26x² + x⁴. If x < 182, the right is greater. If x >= 182, the right swamps x. I couldn't go any further, except that I did notice that 182 = 169 + 13 = 13² + 13, as the top answer begins.
Hi
Can you solve any 4th order equation by this method?
I mean if the coefficient of X is fraction then how do you proceed?
No you can't
Enjoy it
Interesting method.
I did exactly the same as the 1st method. Your videos are fun.
Thank you! ☺️
@@SyberMath thnx for replying :)
x⁴-8x²-x+12=0 can be solved with the first method and yield to four real solution
Loved the video!!
Thank you!!
Geometry problem solve coordinate
Every other day, we will do a geometry puzzle. Thanks for watching!
@@SyberMath I got x equals - square root of n26ntimes 3ndivided byn2..using quadratic formula tosolve for x squared..this is valid too..do you see what I mean?
@@SyberMath Hope you can comment on my alternate solution when you get a chance...thanks..
@@leif1075 I don't understand what you mean by your solution. Can you clarify? Thanks.
@@SyberMath i am stuck at a problem can you help me out.Its,
If the area of an arbitary triangle is 'A'
then find the maximum possible value of its perimeter.
6:01 super nice
Thank you!
Oh, I figured out there were no real roots, and din't think to continue! O.O
nice
how i can solve this ? x^4-3x^2-x+1=0 this is for situation sin and cos and tan and cotg for p/7 radian please guide to me
Problem is that the roots of this equation are expressed with complex radicals so in real numbers they can be expressed with trigonometric functions only
The conclusion is that this quartic will not help with your problem
The equation x+y+1=0 has no solutions because for any x,y x+y+1>=27
im sorry but i didnt get how to solve the sixth degrre equation divided by 4a^2
Sixth degree?
I would say wow 😲
I like 👍 it
Glad to hear that
متد۲محاسبه ًضرایب ومتد۱ جایگزینی وبسیار جالب بود.
Good
Thanks
Both methods are wrong, just verify solution by putting calculated set of x.
Wait!
U assumed that x is real, to take square root on both sides, but then we got non real solutions, ??
You could've also used this method!
ruclips.net/video/j6ri-2S-hxU/видео.html
Unbelievable
WOW 😲
😊
There is error. Two solutions are resl
it will be root of 53
Gorgeous
Thank you!
sorry ot is correct.
Chose a horrible example for polynomial equation 😂 all 4 answers are imaginary