(x^2+x-6)(x^2-3x-4)=24
Find the valie of x
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You need to open the brackets such that:
x^4 + x^3 - 6x^2 - 3x^3 - 3x^2 + 18x + 4x^2 + 4x - 24 = 24
You need to group like terms such that:
x^4 - 2x^3 - 5x^2 + 22x - 48 = 0
You need to notice that x = 3 is a root for the quartic polynomial such that:
x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(ax^3 + bx^2 + cx + d)
Opening the brackets yields:
x^4 - 2x^3 - 5x^2 + 22x - 48 = ax^4 + bx^3 + cx^2 + dx - 3ax^3 - 3bx^2 - 3cx - 3d
Equating the coefficients of like powers yields:
a=1
b-3a=-2 => b = 1
-3d = -48 => d = 16
d-3c=22 => -3c = 6 => c = -2
x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(x^3 + x^2 - 2x + 16)
You need to solve for x the equation x^3 + x^2 - 2x + 16 = 0 such that:
x^3 + x^2 - 2x + 16 = 0
You should come up with the following substitution 3y = 3x + 1 such that:
(y-1/3)^3 + (y-1/3)^2 - 2(y-1/3) + 16 = 0
Expanding the cube and the square yields:
y^3 - y^2 + y/9 - 1/27 + y^2 - 2y/3 + 1/9 - 2y + 2/3 + 16 = 0
Combining like terms yields:
y^3 - 7y/3 + 452/27 = 0
27y^3 - 63y + 452 = 0
Notice that there are no real solutions to equation 27y^3 - 63y + 452 = 0 , hence, the given equation (x^2 -3x+4)(x^2+x-6)=24 has two real solutions, x = 3 and x in (-3,-3.5) and two complex conjugate solutions.
x^4 + x^3 - 6x^2 - 3x^3 - 3x^2 + 18x + 4x^2 + 4x - 24 = 24
You need to group like terms such that:
x^4 - 2x^3 - 5x^2 + 22x - 48 = 0
You need to notice that x = 3 is a root for the quartic polynomial such that:
x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(ax^3 + bx^2 + cx + d)
Opening the brackets yields:
x^4 - 2x^3 - 5x^2 + 22x - 48 = ax^4 + bx^3 + cx^2 + dx - 3ax^3 - 3bx^2 - 3cx - 3d
Equating the coefficients of like powers yields:
a=1
b-3a=-2 => b = 1
-3d = -48 => d = 16
d-3c=22 => -3c = 6 => c = -2
x^4 - 2x^3 - 5x^2 + 22x - 48 = (x-3)(x^3 + x^2 - 2x + 16)
You need to solve for x the equation x^3 + x^2 - 2x + 16 = 0 such that:
x^3 + x^2 - 2x + 16 = 0
You should come up with the following substitution 3y = 3x + 1 such that:
(y-1/3)^3 + (y-1/3)^2 - 2(y-1/3) + 16 = 0
Expanding the cube and the square yields:
y^3 - y^2 + y/9 - 1/27 + y^2 - 2y/3 + 1/9 - 2y + 2/3 + 16 = 0
Combining like terms yields:
y^3 - 7y/3 + 452/27 = 0
27y^3 - 63y + 452 = 0
Notice that there are no real solutions to equation 27y^3 - 63y + 452 = 0 , hence, the given equation (x^2 -3x+4)(x^2+x-6)=24 has two real solutions, x = 3 and x in (-3,-3.5) and two complex conjugate solutions.
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