# Solve the equation x^4 - 5x^2 - 36 = 0

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We have to solve x^4 - 5x^2 - 36 = 0

Let y = x^2

Now x^4 - 5x^2 - 36 = 0

=> y^2 - 5y - 36 = 0

=> y^2 - 9y + 4y - 36 =0

=> y(y -9) + 4(y - 9) = 0

=> (y + 4)(y - 9) = 0

For y + 4 = 0, y = -4

For y - 9 = 0 , y = 9

As x^2 = y, we have x^2 = -4 and 9

=> x = -2i , +2i , -3 an +3.

**Therefore the solutions of the equation x^4 - 5x^2 - 36 = 0 are 2i, -2i , 3 and -3.**

We'll substitute x^2 = t and we'll re-write the equation in t;

t^2 - 5t - 36 = 0

We'll solve the quadratic, using the formula:

t1 = [-b+sqrt(b^2 - 4ac)]/2a

t2 = [-b-sqrt(b^2 - 4ac)]/2a

We'll identify the coefficients a,b,c:

a = 1 , b = -5 , c = -36

t1 = [5+sqrt(25 + 144)]/2

t1 = (5+13)/2

t1 = 9

t2 = (5-13)/2

t2 = -4

But x^2 = t1 => x^2 = 9 => x1 = 3 and x2 = -3

x^2 = t2 => x^2 = -4 => x3 = 2i and x4 = -2i

**The solutions of the equation are: {-3 ; 3 ; -2i ; 2i}.**