Answer:
x = 3, -3, 2i and -2i where i is an imaginary number.
Step-by-step explanation:
We can express the equation \(x^4-5x^2-36=0\) as \((x^2)^2-5x^2-36=0\), and we can make \(y = x^2\). So, we have the equation \(y^2-5y-36=0\); to find the solutions of this last equation, we should find two numbers such that these numbers sum up to -5 and multiplied result in -36, these numbers are -9 and 4. Therefore \(y^2-5y-36=0\) is equivalent to \((y-9)(y+4)=0\), but \(y = x^2\), so, \((x^2-9)(x^2+4)=0\), i.e., \((x+3)(x-3)(x^2+4)=0\). This last product is equal to zero when x = 3, x = -3 or \(x^2=-4\), \(x=\pm\sqrt{-4}\), i.e., \(x=\pm2i\). Therefore the solutions of the equation \(x^4-5x^2-36=0\) are x = 3, -3, 2i and -2i where i is an imaginary number.
