# Solve for real and complex solutions.Integral of 4x^3+6x =10

justaguide | College Teacher | (Level 2) Distinguished Educator

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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}

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

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}.