Respuesta :

isyllus

Answer:

All four roots are complex.

Step-by-step explanation:

Given: [tex]7+5x^4-3x^2[/tex]

First we write as standard form.

[tex]5x^4-3x^2+7[/tex]

Let x² be y

For finding zeros to set the equation 0

[tex]5y^2-3y+7=0[/tex]

Using quadratic formula:

[tex]y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

a=5, b=-3 and c=7

[tex]y=\dfrac{3\pm\sqrt{9-140}}{10}[/tex]

[tex]y=\dfrac{3\pm\sqrt{131}i}{10}[/tex]

[tex]x^2=\dfrac{3\pm\sqrt{131}i}{10}[/tex]

Square root of complex number is complex.

Hence, All four roots are complex.

aksnkj

The given polynomial is a four-degree polynomial. The given polynomial has four roots which are all complex.

Given:

The polynomial is [tex]7+5x^4-3x^2[/tex]

Let [tex]y=x^2[/tex]. Arrange the polynomial as,

[tex]7+5x^4-3x^2=5x^4-3x^2+7\\=5y^2-3y+7[/tex]

Solve the above quadratic polynomial using the quadratic formula,

[tex]y=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=\dfrac{3\pm\sqrt{(-3)^2-4\times 5\times7}}{2\times5}\\=\dfrac{3\pm\sqrt{-131}}{10}\\=\dfrac{3\pm i\sqrt{131}}{10}[/tex]

So, the value of x will be,

[tex]y=x^2=\dfrac{3\pm i\sqrt{131}}{10}\\x=\pm\sqrt {\dfrac{3\pm i\sqrt{131}}{10}}[/tex]

The value of x will be complex number.

Therefore, the given polynomial has four roots which are all complex.

For more details, refer to the link:

https://brainly.com/question/2557258

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