An Interesting Homemade Radical Equation

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[@williamperez-hernandez3968]

Error occured b/c 8^2 + 240 = 304, not 284. 304=16*19, so two solutions include sqrt(19) as part of the answer, not sqrt(17).

[@SyberMath]

I'm not good at arithmetic 😁

[@NadiehFan]

The discriminant of 20x² + 8x − 3 is 8² − 4·20·(−3) = 64 + 240 = 304, not 284.

[@SyberMath]

I'm not good at arithmetic 😁

[@leif1075]

​@@SyberMatheven though rational roots doesnt work you could stil try to break it up.as (ax^2 + bx +c)(dx^2 + fx + g) and solve right..That works a lot pf the times and maybe in this case? Sonwhy didn't you try that just wondering?

[@NadiehFan]

@@leif1075 You clearly have no clue and have not tried this yourself. You introduce six unknowns (a, b, c, d, f, g) but a quartic polynomial has only five coefficients so you would end up with a system of five equations in six variables which has infinitely many solutions. Even if you reduce the number of unknowns to four e.g. by letting a = 5 and d = 1, which is a perfectly reasonable choice for factoring a quartic polynomial with a leading term 5x⁴, you still end up with a _very complicated_ system of four equations in four unknowns (b, c, f, g). To see this, consider the equation 5x⁴ + 2x³ − ⁹⁄₂x² − ³⁄₂x + ⁹⁄₁₆ = 0 which we want to factor as (5x² + bx + c)(x² + fx + g) = 0 Expanding gives 5x⁴ + (b + 5f)x³ + (bf + c + 5g)x² + (bg + cf)x + cg = 0 and equating corresponding coefficients then gives us the system of equations b + 5f = 2 bf + c + 5g = −⁹⁄₂ bg + cf = − ³⁄₂ cg = ⁹⁄₁₆ You will most likely _not_ be able to solve this system, because eliminating all variables except for one will inevitably lead to a sixth degree equation which, in this case, is _not_ a bicubic. If you really want to learn how to deal with quartic equations effectively I suggest you start studying my comments on many of the videos dealing with quartic equations on this channel. SyberMath has learned a lot from me, and so can you.

[@-basicmaths862]

This is a nice example. Solve by substitution 1-x^2=y^2, means x^2+y^2=1 2xy=x^2+x-3/4 x^2+y^2+2xy=x^2+x-3/4+1 (x+y)^2=(x+1/2)^2 x+y=x+1/2 y=1/2, x^2+1/4=1 x^2=3/4, x=±√3/2

[@NadiehFan]

Nice, but (x + y)² = (x + ½)² is equivalent with x + y = x + ½ ⋁ x + y = −x − ½ so you did not find all solutions.

[@-basicmaths862]

​@@NadiehFanI know, but I like to give simple solutions.

[@-basicmaths862]

substitute 1-x^2= y^2, x=±√3/2

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@leif1075]

​@@SyberMathBut don't you agree nonone is going to think to do your second method..whybwould theu after all? Thanks for sharing

[@vaggelissmyrniotis2194]

Found x=+-sqrt(3)/2 or x=-1/5+-sqrt(19)/10.

[@neuralwarp]

I see sqrt(a²-b²) and my mind goes straight to sine and cosine.

[@SyberMath]

Does it always work?

[@taniacsibi6879]

Greșeală e omenească. In rest totul ok

[@LeDevBreton]

3rd solution : trigo

[@SyberMath]

Does it work?

[@LeDevBreton]

@@SyberMath yes but much longer than your solutions. with some cos(pi/3) & sin(pi/6) @ the end

[@NadiehFan]

@@SyberMath Yes, but you will have difficulty expressing all solutions in algebraic form. Since the original equation 2x√(1 − x²) = x² + x − ³⁄₄ implies that any real solution must be on the interval [−1, 1] we could substitute x = sin θ and rewriting the equation as 2x√(1 − x²) = (x + ¹⁄₂)² − 1 this substitution gives 2·sin θ·cos θ = (sin θ + ¹⁄₂)² − 1 provided we assume −¹⁄₂π ≤ θ ≤ ¹⁄₂π in order that cos θ is nonnegative so that √(1 − x²) = cos θ. Note that this constraint does not affect the solutions of the original equation because x = sin θ can take on all values on the interval [−1, 1] if θ ranges over the interval [−¹⁄₂π, ¹⁄₂π]. Adding cos²θ + sin²θ = 1 to both sides of our trigonometric equation we have cos²θ + sin²θ + 2·sin θ·cos θ = (sin θ + ¹⁄₂)² (cos θ + sin θ)² = (sin θ + ¹⁄₂)² cos θ + sin θ = sin θ + ¹⁄₂ ⋁ cos θ + sin θ = −sin θ − ¹⁄₂ cos θ = ¹⁄₂ ⋁ cos θ + 2·sin θ = − ¹⁄₂ Now, the first of these trigonometric equations is easy to solve. Taking θ ∈ [−¹⁄₂π, ¹⁄₂π] into account this gives the solutions θ = ¹⁄₃π ⋁ θ = −¹⁄₃π and therefore x = sin(¹⁄₃π) = ¹⁄₂√3 ⋁ x = sin(−¹⁄₃π) = −¹⁄₂√3. However, the second trigonometric equation cos θ + 2·sin θ = − ¹⁄₂ is not so easy to solve (although there is a standard way to do this). Challenge: solve this equation for θ ∈ [−¹⁄₂π, ¹⁄₂π] and prove that its solution is equivalent with the algebraic solution x = −¹⁄₅ − ¹⁄₁₀√19 of the original equation. Hint: if you solve this trigonometric equation for θ ∈ [−¹⁄₂π, ¹⁄₂π] you should find θ = arcsin(−¹⁄₁₀√5) − arcsin(¹⁄₅√5) so we have x = sin(arcsin(−¹⁄₁₀√5) − arcsin(¹⁄₅√5)) and you should therefore prove that sin(arcsin(−¹⁄₁₀√5) − arcsin(¹⁄₅√5)) = −¹⁄₅ − ¹⁄₁₀√19. To verify this in WolframAlpha, you can enter sin(arcsin(-(1/10)*sqrt(5)) - arcsin((1/5)*sqrt(5))) = -1/5 - (1/10)*sqrt(19) and it will return _True_

[@NadiehFan]

@@LeDevBreton If this is what you got then your trigonometric solution is almost certainly wrong. Can you show your complete solution? Solving the equation trigonometrically I obtained x = sin(¹⁄₃π) ⋁ x = sin(−¹⁄₃π) ⋁ x = sin(arcsin(−¹⁄₁₀√5) − arcsin(¹⁄₅√5)) which equates to x = ¹⁄₂√3 ⋁ x = −¹⁄₂√3 ⋁ x = −¹⁄₅ − ¹⁄₁₀√19

[@LeDevBreton]

@@NadiehFan I may have made a mistake I'll do the calc again tonight and come back

[@Don-Ensley]

problem 2x √(1-x²) = x²+x-3/4 Domain -1

[@leif1075]

Why make x t/2..i dont see why anyone would think of that..why not 4x since 4 squared is 16 or jjst dont sub yet

[@Don-Ensley]

@ it is to make the leading coefficient 1

[@SyberMath]

very nice!

[@pwmiles56]

2x sqrt(1 - x^2) = x^2 + x - 3/4 2x sqrt(1 - x^2) = (x + 1/2)^2 - 1 1 + 2x sqrt(1 - x^2) = (x + 1/2)^2 (x + sqrt(1 - x^2))^2 = (x + 1/2)^2 x + sqrt(1 - x^2) = +/-(x + 1/2) With +ve sign sqrt(1 - x^2) = 1/2 1 - x^2 = 1/4 x^2 = 3/4 x = +/- sqrt(3)/2 With -ve sign sqrt(1 - x^2) = -(2x + 1/2) 1 - x^2 = 4x^2 + 2x + 1/4 5x^2 + 2x - 3/4 = 0 x = (-2 +/-sqrt(4 + 15))/10 x = (-2 +/- sqrt(19)) / 10 The +ve root gives a -ve for the original square root so it doesn't work. The -ve root works.

[@leif1075]

I don't see whybyou would think to move the 1 over and rewrite 1 times x times sqrt as a square like that though..seems too contrived like why would you see that?..Inrewrote the right hand side as (×+3/2)(×-1/2) then tried tonfind common factors..isnt rhat what ppl.will be mire likely to do and more orgabic..and like why didnt you rewrite the riggt hand side that way instead of as x+1/2)^2..not sure why anyonw wouldd see that especially inscr there is no +1/4 there. ..thanks for sharing

[@giuseppemalaguti435]

√(1-x^2)=y...x^2=1-y^2,x=√(1-y^2)...2√(1-y^2)y=1-y^2+√(1-y^2)-3/4=1/4-y^2+√(1-y^2)...√(1-y^2)(2y-1)=(1/2-y)(1/2+y)...2√(1-y^2)(y-1/2)=(1/2-y)(1/2+y)....quindi prime soluzioni sono per y=1/2(x=√3/2,x=-√3/2)...poi semplifico...2√(1-y^2)=-1/2-y...5y^2+y-15/4=0..y=(-1+2√19)/10(no)...y=(-1-2√19)/10,da cui ricavo le altre x,x=-1/5-√19/10..forse ho fatto un po di caos sui segni

[@NadiehFan]

At 2:09 you have arrived at the quartic equation 5x⁴ + 2x³ − ⁹⁄₂x² − ³⁄₂x + ⁹⁄₁₆ = 0 You believe that solving this quartic would be super painful, but it is not and I like a bit of a challenge, so here it goes. First we multiply both sides by 16 to get rid of the fractions, which gives 80x⁴ + 32x³ − 72x² − 24x + 9 = 0 Now, we like to be able to write the leading term as the square of an integer multiple of x², and in order to do so the coefficient of the leading term should be the square of an integer, but 80 = 2⁴·5 is not the square of an integer since it contains an even number of prime factors 2 but only a single prime factor 5. But we can easily see that we can make the coefficient of the leading term into a square of an integer by multiplying both sides of the equation by 5, since we then get a leading coefficient 80·5 = 400 = 2⁴·5² = (2²·5)² = 20². So, we again multiply both sides of the equation, this time by 5, to get 400x⁴ + 160x³ − 360x² − 120x + 45 = 0 Bringing all terms of the second and lower degrees over to the right hand side we get 400x⁴ + 160x³ = 360x² + 120x − 45 At the left hand side 400x⁴ = (20x²)² is the square of 20x² and 160x³ = 2·(20x²)·(4x) is twice the product of 20x² and 4x. So, in accordance with the identity a² + 2·a·b + b² = (a + b)², we can complete the square at the left hand side by adding (4x)² = 16x² to both sides, which gives 400x⁴ + 160x³ + 16x² = 376x² + 120x − 45 which we can write as (20x² + 4x)² = 376x² + 120x − 45 The left hand side is now a perfect square, but the right hand side is not. In order to make the right hand side into a perfect square as well we need to add something to both sides in such a way that the left hand side will _remain_ a perfect square. If we take any number k and add 2k(20x² + 4x) + k² = 40kx² + 8kx + k² to both sides, then the left hand side will remain a perfect square regardless of the value of k. This is because, again in accordance with the identity a² + 2·a·b + b² = (a + b)², we will then have (20x² + 4x)² + 2·(20x² + 4x)²·k + k² = (20x² + 4x + k)² at the left hand side. So, adding 2k(20x² + 4x) + k² = 40kx² + 8kx + k² to both sides we have (20x² + 4x)² + 2k(20x² + 4x) + k² = 376x² + 120x − 45 + 40kx² + 8kx + k² which we can write as (20x² + 4x + k)² = (40k + 376)x² + (8k + 120)x + (k² − 45) Since the left hand side remains a perfect square regardless of the value of k, we are now free to choose k in such a way that the right hand side will also become a perfect square. The right hand side of our equation is a quadratic in x, and a quadratic ax² + bx + c will be a perfect square (that is, the square of a linear polynomial in x _or_ the square of a constant) if and only if its discriminant b² − 4ac is zero. Here we have a = 40k + 376, b = 8k + 120, c = k² − 45, so the right hand side of our equation will be a perfect square if and only if k satisfies (8k + 120)² − 4(40k + 376)(k² − 45) = 0 This is a cubic equation in k, which is generally referred to as the cubic _resolvent_ of our quartic equation. However, quartics with integer coefficients from math contests or home-made problems usually have a nice factorization into two quadratics with integer coefficients, and we can take advantage of this. Assuming our quartic 80x⁴ + 32x³ − 72x² − 24x + 9 = 0 with integer coefficients and therefore also our quartic 400x⁴ + 160x³ − 360x² − 120x + 45 = 0 does indeed have a factorization into two quadratics with integer coefficients we can start by checking if there exists an integer value of k which makes the right hand side (40k + 376)x² + (8k + 120)x + (k² − 45) of our transformed quartic equation the square of a linear polynomial mx + n with integer coefficients m and n. Since (mx + n)² = m²x² + 2mnx + n², we only need to check integer values of k which make the coefficient 40k + 376 = m² of x² the square of an integer m and check if this same integer value of k also makes the constant term k² − 45 = n² the square of an integer n. Moreover, here we only need to check integer values of k which make 40k + 376 the square of an _even_ integer because 40k + 376 cannot be the square of an odd integer for any integer k. First, we try 40k + 376 = 2² = 4 but this does not give an integer value of k. Next, we try 40k + 376 = 4² = 16 which is true for k = −9 and with this value of k the constant term k² − 45 = (−9)² − 45 = 81 − 45 = 36 = 6² is indeed the square of an integer, as required. So, we choose k = −9 and with this value of k our quartic equation (20x² + 4x + k)² = (40k + 376)x² + (8k + 120)x + (k² − 45) becomes (20x² + 4x − 9)² = 16x² + 48x + 36 which can be written as (20x² + 4x − 9)² = (4x + 6)² Bringing over the square from the right hand side to the left hand side this gives (20x² + 4x − 9)² − (4x + 6)² = 0 and applying the difference of two squares identity a² − b² = (a − b)(a + b) to the left hand side this gives (20x² − 15)(20x² + 8x − 3) = 0 Note that the first quadratic factor is 5 times a quadratic with integer coefficients. This is quite understandable because we multiplied both sides of 80x⁴ + 32x³ − 72x² − 24x + 9 = 0 by 5 to obtain 400x⁴ + 160x³ − 360x² − 120x + 45 = 0. Dividing both sides by 5 we get (4x² − 3)(20x² + 8x − 3) = 0 which is therefore an integer factorization of our quartic 80x⁴ + 32x³ − 72x² − 24x + 9 = 0. Applying the zero product property A·B = 0 ⟺ A = 0 ⋁ B = 0 this gives 4x² − 3 = 0 ⋁ 20x² + 8x − 3 = 0 and solving these two quadratics then gives x = ¹⁄₂√3 ⋁ x = −¹⁄₂√3 ⋁ x = −¹⁄₅ + ¹⁄₁₀√19 ⋁ x = −¹⁄₅ − ¹⁄₁₀√19 Of course, we need to validate each of these solutions as solutions of the original equation 2x√(1 − x²) = x² + x − ³⁄₄ because the quartic equation we solved was obtained by squaring both sides of the original equation which implies that the solutions of the quartic equation also include any solution of the equation 2x√(1 − x²) = −(x² + x − ³⁄₄). As we can see any valid real solution of the original equation should be on the closed interval [−1, 1] since 1 − x² should not be negative, but all four solutions of the quartic satisfy this criterion. However, this does _not_ mean that all four solutions of the quartic are valid solutions of the original equation. The right hand side x² + x − ³⁄₄ = (x + ¹⁄₂)² − 1 is negative for x ∈ (−³⁄₂, ¹⁄₂) so any valid positive solution of the original equation should be on the interval [¹⁄₂, 1] while any valid negative solution can be on the interval [−1, 0]. As a consequence, only the solution x = −¹⁄₅ + ¹⁄₁₀√19 must be discarded as a solution of the original equation since we have 4 < √19 < 5 and therefore ²⁄₅ < ¹⁄₁₀√19 < ¹⁄₂ and therefore ¹⁄₅ < −¹⁄₅ + ¹⁄₁₀√19 < ³⁄₁₀. It is easy to check that, conversely, x = −¹⁄₅ + ¹⁄₁₀√19 is indeed the sole solution of the equation 2x√(1 − x²) = −(x² + x − ³⁄₄).

[@StaR-uw3dc]

Nice. Would the substitution t=2x in the equation 80x⁴ + 32x³ − 72x² − 24x + 9 = 0 lead to easier calculations (5t⁴ + 4t³ − 18t² − 12t + 9 = 0)?

[@SyberMath]

You're writing a book with your comments! 😍❤️

[@NadiehFan]

@@StaR-uw3dc Actually, no. To solve the equation 5t⁴ + 4t³ − 18t² − 12t + 9 = 0 using Ferrari's method (which is the method I used) we first make the coefficient of the leading term the square of an integer by multiplying both sides by 5, which gives 25t⁴ + 20t³ − 90t² − 60t + 45 = 0 The coefficients are smaller (except for the constant term), but following the same procedure you will get the exact same cubic resolvent with integer solution k = −9 and you will end up with (5t² + 2t − 9)² = (2t + 6)² Substituting t = 2x this gives (20x² + 4x − 9)² = (4x + 6)² which is of course exactly what I got from 400x⁴ + 160x³ − 360x² − 120x + 45 = 0.

[@StaR-uw3dc]

@@NadiehFan Thank you.

[@SyberMath]

Wow!!!