Respuesta :

alinakincsem

Answer:

x = ±5

Step-by-step explanation:

We are given the following polynomial function and we are to find all of its real roots:

[tex]x^4-24x^2-25[/tex]

Let [tex]y=x^2[/tex] so we can now write it as:

[tex]y^2-24y-25[/tex]

Factorizing it to get:

[tex](y^2+y)+(-25y-25)[/tex]

[tex]y\left(y+1\right)-25\left(y+1\right)[/tex]

[tex] \left ( y + 1 \right ) \left ( y - 2 5 \right)[/tex]

Substitute back [tex]y=x^2[/tex] to get:

[tex] \left ( x^2 + 1 \right ) \left ( x^2 - 25 \right ) [/tex]

[tex] \left ( x^2 + 1 \right ) \left ( x + 5 \right ) \left ( x - 5 \right ) [/tex]

The quadratic factor has only complex roots. Therefore, the real roots are x = ±5.

imlittlestar

Answer:

The real roots are  ±5

Step-by-step explanation:

It is given that,

f(x) = x^4 - 24x^2 - 25

To find the real roots

Let x^4 - 24x^2 - 25 = 0 ----(1)

Take y = x^2

Then eq (1) becomes,

y^2 - 24y - 25 = 0

By using splitting method we can write,

y^2 + y - 25y - 25 = 0

y(y + 1) - 25(y +1) = 0

(y + 1)(y - 25) = 0

(x^2 + 1 )(x^2 - 25) = 0

From (x^2 + 1 ) we get complex roots

x^2 = -1

x = √-1

x - 25 = 0 we get real roots

x = 25

x = ±5

Therefore the real roots are  ±5