Answer:
C
Step-by-step explanation:
We are given the function:
[tex]\displaystyle f(x) = x^4 - 8x^2 + 10[/tex]
And we want to determine between which consecutive integers are the real zeros of f located.
Because the equation is in quadratic form, we can use the quadratic formula. First, we can let u = x². Then:
[tex]\displaystyle f(u) = u^2 - 8u + 10 = 0[/tex]
Solve using the quadratic formula:
[tex]\displaystyle \begin{aligned} u &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ & = \frac{-(-8)\pm\sqrt{(-8)^2 -4(1)(10)}}{2(1)} \\ \\ & = \frac{8\pm\sqrt{24}}{2} \\ \\ &=\frac{8\pm2\sqrt{6}}{2}} \\ \\ & = 4\pm\sqrt{6} \end{aligned}[/tex]
Hence:
[tex]\displaystyle u = 4+\sqrt{6} \text{ or } u = 4-\sqrt{6}[/tex]
Back-substitute:
[tex]x^2 = 4+\sqrt{6} \text{ or } x^2 = 4-\sqrt{6}[/tex]
Solve for x:
[tex]\displaystyle x = \pm\sqrt{4+\sqrt{6}} \text{ or } x=\pm\sqrt{4-\sqrt{6}}[/tex]
Hence, the four real solutions of x are:
[tex]\displaytstyle x = -\sqrt{4-\sqrt{6}}, -\sqrt{4+\sqrt{6}}, \sqrt{4-\sqrt{6}}, \sqrt{4+\sqrt{6}}[/tex]
Or, approximately:
[tex]\displaystyle x \approx -1.245, -2.540, 1.245, 2.540[/tex]
Respectively.
Therefore, our answers are both A and B, or simply C.
