A Very Nice Math Olympiad Problem | Solve for the values of x? | Algebra Equation

[@kareolaussen819]

At about 4:30 the lecturer is cheating: The tentative factorization over the integers must take the form (x^2+rx+p)(x^2-rx+q) There is no stated prior reason why r should equal 1 (except by already knowing the result)! With r unknown a completely different solution strategy must be adopted, based on all possible factorizations of the constant term.

[@1234larry1]

I tried x^4-14x+5x+30=0, using your method, but the system of equations was q-p=5, q+p=-13 and qp=30. Two of the solutions involved quadratics with “c” that contained a square root for a coefficient.

[@nicoz5787]

I suspect you could try to decompose your polynomial as a product of two factors (x²+a*x+p)*(x²-a*x+q) and thus solve a system of 3 equations with 3 unknowns.

[@raghvendrasingh1289]

we will use Descartes' method for solving biquadratic equation Let x^4 - 14 x^2 +5x +30 = (x^2 +ax+p)(x^2-ax+q) p q = 30 a(q-p) = 5 p+q - a^2 = - 14 now (q+p)^2 - (q - p)^2 = 4qp hence (a^2- 14)^2 - 25/a^2 = 120 we put t = a^2 t^2 - 28 t+196-25/t-120= 0 t^3 - 28 t^2+76 t- 25 = 0 t = a^2 hence we will use RRT with t = 1 , 25 t = 25 satisfies the equation hence a = 5 q+p = 11 , q - p = 1 q = 6 , p = 5 (x^2+5x+5)(x^2 - 5x+6) however in this case we can factorise easily by RRT by taking x = 2 , 3

[@kareolaussen819]

Introduce y=√(x+5), and rewrite equation as x^2 - y = 5 y^2 - x = 5 Subtract to find x^2 -y^2+x-y=(x-y)(x+y+1)=0. Case y=x => (after squaring) (x-1/2)^2 = 5+1/4 = 21/4 x=(1+√21)/2 is a valid solution, x=(1-√21)/2 is not a valid expression for y as a square root. Case y=-(x+1) => (after squaring) (x+1/2)^2 = 5-1+1/4 = 17/4 x=-(1+√17)/2 is a valid solution. x=(-1+√17)/2 does not lead to a valid expression for y as a square root.

[@mercedesclaveras]

La equis subuno en positiva mayor que cero la que es negativa es la equis subidos

[@nguyenthang946]

How do we know a quartic equation without a cubic can be factorised as (x²+x+p)(x²-x+q) and does it still hold true even when there is a cubic? Also is there any reasoning why the coefficient of x equals 1 anyway? 😄

[@1234larry1]

It’s tricky, but not as hard as partial fractions.

[@ジャスミン-v1s]

为什么能展开成这两个二次方程的乘积 我随便写了个方程，就无法这样做 x^4 -22x^2+3x+8=0

[@RealQinnMalloryu4]

(x^2)^2 ➖(5)^2={x^4 ➖ 25}=21 3^7 3^7^1 3^1^1 3^1 (x ➖ 3x+1). {x+x ➖ }+{5+5 ➖ }={x^2+10}=10x^2 5^5x^2 2^3^2^3x^2 1^1^1^3x^2 3x^2 (x ➖ 3x+2).