Question

What are the real or imaginary solutions of the polynomial equation x^4-52x^2+576?

Answer 1

The roots of the polynomial for this case will be x1 = 6, x2 = -6, x3 = 4, x4 = -4.You can solve the polynomial using the resolver after making the variable change u = x ^ 2 so that you have a second degree polynomial. Then return the change to find the missing roots. Attached solution.

Answer 2

Answer:

$x=-6,-4,4,6$.

Step-by-step explanation:

We have been given an equation $x^4-52x^2+576=0$. We are asked to find the real or imaginary solutions of the given equation.

First of all, we will substitute $u=x^2$ in our given equation as:

$(x^2)^2-52x^2+576=0$

$(u)^2-52u+576=0$    

Now, we  will solve for u as:

$u^2-36-16u+576=0$

$u(u-36)-16(u-36)=0$

$(u-36)(u-16)=0$

$u-36=0\text{ (or) }u-16=0$

$u=36\text{ (or) }u=16$  

Now, we will undo substitution as:

$x^2=36\text{ (or) }x^2=16$

$x=\pm\sqrt{36}\text{ (or) }x=\pm\sqrt{16}$

$x=\pm6\text{ (or) }x=\pm\sqrt4$

Therefore, the solutions of our given equation are $x=-6,-4,4,6$.