Question

Please answer correctly !!!!!!! Will mark Brianliest !!!!!!!!!!!

Answer 1

Answer:

$x=-2\\x=2$

(You only need to give one solution)

Step-by-step explanation:

We have the following equation

$(x^2+1)^2-5x^2-5=0$

First, we need to foil out the parenthesis

$x^4+2x^2+1-5x^2-5=0$

Now we can combine the like terms

$x^4-3x^2-4=0$

Now, we need to factor this equation.

To factor this, we need to find a set of numbers that add together to get -3 and multiply to give us -4.

The pair of numbers that would do this would be 1 and -4.

This means that our factored form would be

$(x^2-4)(x^2+1)=0$

As the first binomial is a difference of squares, it can be factored futher into

$(x^2+1)(x+2)(x-2)=0$

Now, we can get our solutions.

The first binomial will produce two complex (Not real) solutions.

$x+2=0\\\\x=-2$

$x-2=0\\\\x=2$

So our solutions to this equation are

$x=-2\\x=2$