Since that factoring was beyond me, I split the equation into two quadratics: (x^2+ax+5) and (x^2-bx-1) The object after this is to multiply the two “dummy” quadratics, combine the coefficients of x^3, combine the coefficients of x^2, and combine the coefficients of x to get a system of equations using a and b. Then match the coefficients of x^3, x^2, and x with the corresponding coefficients in the original equation. Here you have a system of equations you can use to solve for a and b. When you find a and b, substitute the values for a and b for the middle terms in the “dummy” quadratics, then solve each quadratic.
@@musicsubicandcebu1774yes, you have to keep trying until you get the correct factors and signs on each side, for instance, when I started out I had (x^2-ax-5)(x^+bx+1) and the numbers didn’t work until I put (x^2+ax+5)(x^2-bx-1). Now this is tricky because it turned out that a=b, but since I already had -ax and +bx set up, when I found out both a and b=2 respectively, putting 2 into both factors changed the sign automatically into 2 and -2.
Since that factoring was beyond me, I split the equation into two quadratics: (x^2+ax+5) and (x^2-bx-1)
The object after this is to multiply the two “dummy” quadratics, combine the coefficients of x^3, combine the coefficients of x^2, and combine the coefficients of x to get a system of equations using a and b. Then match the coefficients of x^3, x^2, and x with the corresponding coefficients in the original equation. Here you have a system of equations you can use to solve for a and b. When you find a and b, substitute the values for a and b for the middle terms in the “dummy” quadratics, then solve each quadratic.
Very nice! ❤
Easier said than done!
@@musicsubicandcebu1774yes, you have to keep trying until you get the correct factors and signs on each side, for instance, when I started out I had (x^2-ax-5)(x^+bx+1) and the numbers didn’t work until I put (x^2+ax+5)(x^2-bx-1). Now this is tricky because it turned out that a=b, but since I already had -ax and +bx set up, when I found out both a and b=2 respectively, putting 2 into both factors changed the sign automatically into 2 and -2.
👍
x^4 - 12 x - 5 = 0
x^4+4 - 12 x - 9 = 0
(x^2+2)^2 - (4 x^2+12 x+9) = 0
(x^2+2)^2 - (2x+3)^2 = 0
(x^2+2x+5) (x^2 - 2x - 1) = 0
case 1
x^2+2x = - 5
(x+1)^2 = - 4
x = - 1+2 i , - 1 - 2 i
case 2
x^2 - 2x = 1
(x - 1)^2 = 2
x = 1+√2 , 1 - √2
Far better than upper example ,very good 👍.
Very nice! ❤