Solve for real and complex solutions. Integral of 4x^3+6x =10
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We have Int[ 4x^4 + 6x dx] = 10
=> 4*x^4 / 4 + 6x^2 / 2 = 10
=> x^4 + 3x^2 = 10
Let y = x^2
=> y^2 + 3y - 10 = 0
y1 = -3/2 + sqrt (49)/2
=> y1 = -3/2 + 7/2
=> y1 = 2
y2 = -3/2 - 7/2
=> y2 = -5
Now y = x^2
So, we have:
x1 = sqrt 2
x2 = -sqrt 2
x3 = i*sqrt 5
x4 = -i*sqrt 5
The real and complex roots are { sqrt 2, -sqrt 2, i*sqrt 5, -i*sqrt 5}
To determine all real and complex solutions of the equation, we'll have to determine the indefinite integral from the left side, for the beginning.
Int (4x^3+6x)dx = 4Int x^3 dx + 6Int x dx
Int (4x^3+6x)dx = 4x^4/4 + 6x^2/2 + C
Int (4x^3+6x)dx = x^4 + 3x^2
We'll substitute the integral from the left side by it's result:
x^4 + 3x^2 = 10
We'll subtract 10 both sides:
x^4 + 3x^2 - 10 = 0
It is a bi-quadratic equation. We'll put x^2 = t.
t^2 + 3t - 10 = 0
We'll apply quadratic formula:
t1 = [-3+sqrt(9+40)]/2
t1 = (-3+7)/2
t1 = 2
t2 = (-3-7)/2
t2 = -5
x^2 = t1 => x^2 = 2
We'll take square root both sides:
x1 = -sqrt2
x2 = sqrt2
x^2 = t2 => x^2 = -5
We'll take square root both sides:
x3 = -isqrt5
x4 = isqrt5
The real and complex roots of the given equation are; {-sqrt2 ; sqrt2 ; -isqrt5 ; isqrt5}.
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