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Determine between which consecutive intergers the real zeros of f(x)= x^4-8x^2+10 are located.

Determine between which consecutive intergers the real zeros of f(x)= x^4-8x^2+10 are-example-1

1 Answer

5 votes

Answer:

C

Explanation:

We are given the function:


\displaystyle f(x) = x^4 - 8x^2 + 10

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:


\displaystyle f(u) = u^2 - 8u + 10 = 0

Solve using the quadratic formula:


\displaystyle \begin{aligned} u &= (-b\pm√(b^2-4ac))/(2a) \\ \\ & = (-(-8)\pm√((-8)^2 -4(1)(10)))/(2(1)) \\ \\ & = (8\pm√(24))/(2) \\ \\ &=(8\pm2√(6))/(2)} \\ \\ & = 4\pm√(6) \end{aligned}

Hence:

\displaystyle u = 4+√(6) \text{ or } u = 4-√(6)

Back-substitute:


x^2 = 4+√(6) \text{ or } x^2 = 4-√(6)

Solve for x:


\displaystyle x = \pm\sqrt{4+√(6)} \text{ or } x=\pm\sqrt{4-√(6)}

Hence, the four real solutions of x are:

\displaytstyle x = -\sqrt{4-√(6)}, -\sqrt{4+√(6)}, \sqrt{4-√(6)}, \sqrt{4+√(6)}

Or, approximately:


\displaystyle x \approx -1.245, -2.540, 1.245, 2.540

Respectively.

Therefore, our answers are both A and B, or simply C.

User Nick Gerner
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