The real roots of a function are the rational and the irrational roots of the function
The real roots are: [tex]\mathbf{x = -2}[/tex], [tex]\mathbf{x = \frac 14}[/tex], [tex]\mathbf{x = -3 - 2\sqrt{5}}[/tex] and [tex]\mathbf{x = -3 + 2\sqrt{5}}[/tex]
The equation is given as:
[tex]\mathbf{ 4x^4+31x^3-4x^2-89x+22=0}[/tex]
Factorize
[tex]\mathbf{(x + 2) (4 x - 1) (x^2 + 6 x - 11) = 0}[/tex]
Split
[tex]\mathbf{x + 2 = 0\ or\ 4 x - 1 = 0 \ or\ x^2 + 6 x - 11 = 0}[/tex]
Solve for x
[tex]\mathbf{x = -2\ or\ x = \frac 14 \ or\ x^2 + 6 x - 11 = 0}[/tex]
Solve
[tex]\mathbf{x^2 + 6 x - 11 = 0}[/tex] using the following quadratic formula:
[tex]\mathbf{x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}[/tex]
So, we have:
[tex]\mathbf{x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times -11}}{2 \times 1}}[/tex]
[tex]\mathbf{x = \frac{-6 \pm \sqrt{80}}{2}}[/tex]
[tex]\mathbf{x = \frac{-6 \pm 4\sqrt{5}}{2}}[/tex]
[tex]\mathbf{x = -3 \pm 2\sqrt{5}}[/tex]
Hence, the real roots are:
[tex]\mathbf{x = -2}[/tex], [tex]\mathbf{x = \frac 14}[/tex], [tex]\mathbf{x = -3 - 2\sqrt{5}}[/tex] and [tex]\mathbf{x = -3 + 2\sqrt{5}}[/tex]
Read more about real roots at:
https://brainly.com/question/21664715
